22.2 - Derivative Test For Increasing And Decreasing functions

In the previous section, we saw definition 1 and definition 2, which help us to find the nature of a function at any given point. In this section, we will see how derivatives can be applied to find the nature.

To apply derivatives, we need to learn theorem 1. We will first see the theorem. Then we will write an explanation and after that, we will write the proof.

Theorem 1
Let f be continuous in [a,b] and differentiable in (a,b). Then
(a) f is strictly increasing in [a,b] if f'(x) > 0 for each x ∈ (a,b)
(b) f is increasing in [a,b] if f'(x) ≥ 0 for each x ∈ (a,b)
(c) f is strictly decreasing in [a,b] if f'(x) < 0 for each x ∈ (a,b)
(d) f is decreasing in [a,b] if f'(x) ≤ 0 for each x ∈ (a,b)
(e) f is constant in [a,b] if f'(x) = 0 for each x ∈ (a,b)

Explanation can be written in 2 steps:
1. We want to know the nature of the function in the interval [a,b]
2. There are infinite number of points in the interval (a,b).
(a) If the derivative f'(x) is greater than zero at all those points, then we say that:
f is strictly increasing in [a,b].

(b) If the derivative f'(x) is greater than or equal to zero at all those points, then we say that:
f is increasing in [a,b].

(c) If the derivative f'(x) is lesser than zero at all those points, then we say that:
f is strictly decreasing in [a,b] 

(d) If the derivative f'(x) is lesser than or equal to zero at all those points, then we say that:
f is decreasing in [a,b]

(e) If the derivative f'(x) is zero at all those points, then we say that:
f is constant in [a,b]

Proof for (a) can be written in 4 steps:
1. Let x₁ and x₂ be any two points in the interval [a,b], such that x₁ < x₂.
2. Given that:
   ♦ f is a real valued function
   ♦ f is continuous in [a,b]
   ♦ f is differentiable in (a,b)
• Then the three conditions are satisfied. We can apply Mean Value Theorem between the two points x₁ and x₂.

3. There exists a point c between x₁ and x₂ in such a way that:
Derivative at c is equal to the slope of the secant between x₁ and x₂.
• So we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(c)}    & {~=~}    &{\frac{f(x_2) \,-\, f(x_1)}{x_2 \,-\, x_1}}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(x_2) \,-\, f(x_1)}    & {~=~}    &{f'(c) (x_2 \,-\, x_1)}    \\
{~\color{magenta}    3    }    &{\implies}    &{f(x_2) \,-\, f(x_1)}    & {~>~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{f(x_2)}    & {~>~}    &{f(x_1)}    \\
\end{array}$

◼ Remarks:
2 (magenta color):
• In part (a) of the theorem, it is given that, f'(x) > 0 for all points within (a,b). So f'(c) will be greater than zero.
• Also, we choose x₁ and x₂ in such a way that x₂ > x₁. So (x₂ − x₁) will be greater than zero.
• Therefore, f'(c) (x₂ − x₁) will be greater than zero.

4. From the above step, we see that:
If x₁ < x₂, then f(x₁) < f(x₂) and so, the given function is a strictly increasing function.   

Proof for (b) can be written in 4 steps:
1. Let x₁ and x₂ be any two points in the interval [a,b], such that x₁ < x₂.
2. Given that:
   ♦ f is a real valued function
   ♦ f is continuous in [a,b]
   ♦ f is differentiable in (a,b)
• Then the three conditions are satisfied. We can apply Mean Value Theorem between the two points x₁ and x₂.

3. There exists a point c between x₁ and x₂ in such a way that:
Derivative at c is equal to the slope of the secant between x₁ and x₂.
• So we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(c)}    & {~=~}    &{\frac{f(x_2) \,-\, f(x_1)}{x_2 \,-\, x_1}}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(x_2) \,-\, f(x_1)}    & {~=~}    &{f'(c) (x_2 \,-\, x_1)}    \\
{~\color{magenta}    3    }    &{\implies}    &{f(x_2) \,-\, f(x_1)}    & {~\ge~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{f(x_2)}    & {~\ge~}    &{f(x_1)}    \\
\end{array}$

◼ Remarks:
2 (magenta color):
• In part (b) of the theorem, it is given that, f'(x) ≥ 0 for all points within (a,b). So f'(c) will be greater than or equal to zero.
• Also, we choose x₁ and x₂ in such a way that x₂ > x₁. So (x₂ − x₁) will be greater than zero.
• Therefore, f'(c) (x₂ − x₁) will be greater than or equal to zero.

4. From the above step, we see that:
If x₁ < x₂, then f(x₁) ≤ f(x₂) and so, the given function is an increasing function.

Proof for (c) can be written in 4 steps:
1. Let x₁ and x₂ be any two points in the interval [a,b], such that x₁ < x₂.
2. Given that:
   ♦ f is a real valued function
   ♦ f is continuous in [a,b]
   ♦ f is differentiable in (a,b)
• Then the three conditions are satisfied. We can apply Mean Value Theorem between the two points x₁ and x₂.

3. There exists a point c between x₁ and x₂ in such a way that:
Derivative at c is equal to the slope of the secant between x₁ and x₂.
• So we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(c)}    & {~=~}    &{\frac{f(x_2) \,-\, f(x_1)}{x_2 \,-\, x_1}}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(x_2) \,-\, f(x_1)}    & {~=~}    &{f'(c) (x_2 \,-\, x_1)}    \\
{~\color{magenta}    3    }    &{\implies}    &{f(x_2) \,-\, f(x_1)}    & {~<~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{f(x_2)}    & {~<~}    &{f(x_1)}    \\
\end{array}$

◼ Remarks:
2 (magenta color):
• In part (c) of the theorem, it is given that, f'(x) < 0 for all points within (a,b). So f'(c) will be less than zero.
• Also, we choose x₁ and x₂ in such a way that x₂ > x₁. So (x₂ − x₁) will be greater than zero.
• Therefore, f'(c) (x₂ − x₁) will be less than zero.

4. From the above step, we see that:
If x₁ < x₂, then f(x₁) > f(x₂) and so, the given function is a strictly decreasing function.

• Proofs for parts (d) and (e) can be written in a similar way.

---

Now we will see some solved examples:

Solved example 22.8
Show that the function given by
f(x) = x³ − 3x² + 4x, x ∈ R is strictly increasing on R.
Solution:
1. f'(x) = 3x² −6x + 4
2. This can be written as:
f'(x) = 3(x² −2x + 1) + 1
= 3(x −1)² + 1
3. There is a square term and a +ve term. So f'(x) will be always greater than zero.
• We can try input x from any interval (a,b) in R. The value of f'(x) will be never less than zero.
• Also, since there is a "+ 1" term, f'(x) can never become equal to zero.
Therefore, f is a strictly increasing function.

4. Fig.22.45 below shows the graph of f(x) in red color and f'(x) in yellow color.

Fig.22.4

From the yellow curve we see that, no f'(x) value is zero or less than zero.

---

In the next section, we will see a few more solved examples.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

Chapter 22.1 - Increasing And Decreasing Functions

In the previous section, we saw how derivatives can be used to find rate of change of quantities. In this section, we will see increasing and decreasing functions.

Some basic details can be written in 4 steps:
1. Consider the graph of the function f(x) = x². It is shown in fig.22.3 below:

Fig.22.3

2. We see a vertical yellow line. This line helps us to understand the concept of height of graph.
• We see that, the vertical yellow line is drawn through x₀.
   ♦ The lower end of the line is on the x-axis.
   ♦ The upper end of the line lies on the curve.
• So height of the vertical yellow line is f(x₀).
• In such a situation, we say that:
Height of the graph at x₀ is f(x₀).

---

• It may be noted that, we do not consider the actual height of the yellow line. We consider the value of f(x₀).

• Consider two cases:
   ♦ Case I: f(x₀) = −8 
   ♦ Case II: f(x₀) = −3
• Height of the yellow line will be larger in case I. But the value of f(x₀) is larger in case II.

---

3. Now consider the portion of the x-axis from O to X'.
• In this portion, if we move along the x-axis from left to right, we will be passing through points like −3, −2, −1.5, −1, −0.5, −0.25, and finally 0.
• We can find the corresponding f(x) values. They are tabulated in the table 22.1(a) below:

Table 22.1(a)

• The second column of this table, gives the corresponding heights. We see that:
As we move from left to right, the height of the graph decreases.
• In this situation, we write:
The function is decreasing for x < 0

4. Similarly, consider the portion of the x-axis from O to X.
• In this portion, if we move along the x-axis from left to right, we will be passing through points like 0, 0.25, 0.5, 1, 1.5, 2, 3, . . .
• We can find the corresponding f(x) values. They are tabulated in the table 22.1(b) above.
• The second column of this table, gives the corresponding heights. We see that:
As we move from left to right, the height of the graph increases.
• In this situation, we write:
The function is increasing for x > 0

---

Based on the above discussion, we can write a definition of increasing and decreasing functions.

Definition 1:

Let I be an open interval contained in the domain of a real valued function f. Then f is said to be
(i) Increasing on I if x₁ < x₂ in I ⇒ f(x₁) ≤ f(x₂) for all x₁, x₂ ∈ I.
(ii) Strictly increasing on I if x₁ < x₂ in I ⇒ f(x₁) < f(x₂) for all x₁, x₂ ∈ I.
(iii) Decreasing on I if x₁ < x₂ in I ⇒ f(x₁) ≥ f(x₂) for all x₁, x₂ ∈ I.
(iv) Strictly decreasing on I if x₁ < x₂ in I ⇒ f(x₁) > f(x₂) for all x₁, x₂ ∈ I.

---

Now we will write a detailed explanation of the definition. The explanation can be written by using the function f(x) = x² as an example. It can be written in 6 steps:

1. In the definition, there is a mention about domain of a real valued function.
• In fig.22.1 above, f(x) = x² is a real valued function. It's domain is R. So all points in the x-axis, are included in the domain.
2. In the definition, there is a mention about an open interval I.
• 'Open interval' means, the end points should not be used as input values.
• In our discussion based on fig.22.1 above, we used two open intervals:
   ♦ (−∞,0) related to the portion OX' of the x-axis.
   ♦ (0,∞) related to the portion OX of the x-axis.
• Both of the above intervals are contained in R.

3. Now consider the open interval (−∞,0).
• Take any two points x₁ and x₂ such that, x₁ < x₂.
Let us take x₁ = −3 and x₂ = −0.25
• So we get:
   ♦ f(x₁) = f(−3) = (−3)² = 9  
   ♦ f(x₂) = f(−0.25) = (−0.25)² = 0.0625
• We can write: f(x₁) > f(x₂)
• So item (iv) of the definition is applicable here. We can write:
On the open interval I = (−∞,0), the function f(x) = x² is strictly decreasing.
• The general graphical representation of a strictly decreasing function is shown in fig.22.4(d) below:

Fig.22.4

• Note that x₁, that we select as the first point, should be always less than x₂, that we select as the second point. This condition will ensure that, we always move from left to right along the x-axis. 

4. Similarly consider the open interval (0,∞).
• Take any two points x₁ and x₂ such that, x₁ < x₂.
Let us take x₁ = 0.5 and x₂ = 3
• So we get:
   ♦ f(x₁) = f(0.5) = (0.5)² = 0.25 
   ♦ f(x₂) = f(3) = (3)² = 9
• We can write: f(x₁) < f(x₂)
• So item (ii) of the definition is applicable here. We can write:
On the open interval I = (0,∞), the function f(x) = x² is strictly increasing.
• The general graphical representation of a strictly increasing function is shown in fig.22.4(b) above.
• Note that x₁, that we select as the first point, should be always less than x₂, that we select as the second point. This condition will ensure that, we always move from left to right along the x-axis.

5. Once we understand strictly decreasing function, we will be able to write about decreasing function. Item (iii) of the definition is applicable here. The only difference between the two can be written as:
   ♦ For strictly decreasing, we have f(x₁) > f(x₂)
   ♦ For decreasing, we have f(x₁) ≥ f(x₂)
• That means, for some x₁ and x₂ values, the corresponding f(x) values may be equal. Such x values will give a horizontal segment to the graph.
• The general graphical representation of a decreasing function is shown in fig.22.4(c) above.

6. Similarly, once we understand strictly increasing function, we will be able to write about increasing function. Item (i) of the definition is applicable here. The only difference between the two can be written as:
   ♦ For strictly increasing, we have f(x₁) < f(x₂)
   ♦ For increasing, we have f(x₁) ≤ f(x₂)
• That means, for some x₁ and x₂ values, the corresponding f(x) values may be equal. Such x values will give a horizontal segment to the graph.
• The general graphical representation of a increasing function is shown in fig.22.4(a) above.

---

Now we will see a solved example.
Solved example 22.7
Show that the function f(x) = 7x − 3 is strictly increasing on R.
Solution:
1. In the discussion above, we considered an interval I inside the domain. In this problem, the interval is same as the domain, which is R.
2. Take any two points x₁ and x₂ from R. x₁ must be less than x₂.
• Let us take x₁ = −3 and x₂ = 5
3. So we get:
   ♦ f(x₁) = 7(−3) − 3 = −24
   ♦ f(x₂) = 7(5) − 3 = 32
4. We see that, f(x₁) < f(x₂).
So based on definition 1, we get a hint that, the given f(x) is strictly increasing. However, to prove it beyond doubt, we must write the proof for the general case. It can be done in 3 steps:

1. Take any two points x₁ and x₂ from R such that, x₁ < x₂.
2. Since x₁ < x₂, we get: 7(x₁) − 3 < 7(x₂) − 3
3. Since, 7(x₁) − 3 < 7(x₂) − 3, we can write: f(x₁) < f(x₂)
• So based on definition 1, we can write: f is a strictly decreasing function.

---

• The definition 1 that we saw above, will help us to find the nature of a function within an interval I.
• The next definition 2 that we are going to see, will help us to find the nature of a function at a point.

We will write the definition first and then see the explanation.
Definition 2:
Let x₀ be a point in the domain of a real valued function f. Then f is said to be increasing, strictly increasing, decreasing or strictly decreasing at x₀ if there exist an open interval I containing x₀ such that f is increasing, strictly increasing, decreasing or strictly decreasing, respectively in I.
The explanation can be written in 3 steps:
1. We are given a real valued function f.
2. x₀ is any point in the domain of f.
We want to know the nature of f at the point x₀.
3. For that, we consider an open interval:
I = (x₀ − h, x₀ + h), where h > 0.
• If I exists such that, f is increasing on I, then we say that f is increasing at x₀.
• If I exists such that, f is strictly increasing on I, then we say that f is strictly increasing at x₀.
• If I exists such that, f is decreasing on I, then we say that f is decreasing at x₀.
• If I exists such that, f is decreasing on I, then we say that f is strictly decreasing at x₀.

---

In the next section, we will see how to apply derivatives to find the nature.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

Chapter 22 - Application of Derivatives

In the previous section, we completed a discussion on continuity and differentiability. We saw how to find the derivatives of various functions.

Derivatives have many applications in science, engineering, economics etc., In this chapter, we will see some of those applications.

First we will see how derivatives help us to find the rate of change of quantities. Let us see an example. It can be written in 6 steps:
1. An object is moving in a straight line.
• It started with an initial velocity zero. That is, it started from rest.
• But it has a constant acceleration 'a'.
• A stop-watch is turned on at the instant when the object starts it's motion.
2. In such a situation, the distance traveled by the object when the stop-watch reading is 't', can be obtained using the formula: $\rm{s \,=\, \frac{1}{2} a t^2   }$
3. Using this formula, we can plot a graph with t along the x-axis and s along the y-axis. If the acceleration 'a' is $\rm{3 ~ m s^{-2}}$, the graph will be as shown in fig.22.1 below:

Fig.22.1

4. Since the object is moving with a constant acceleration 'a', the velocity will be changing continuously. That is, velocity is not uniform.
5. We can find the velocity at any instant by using the derivative $\rm{\frac{ds}{dt}}$ (recall that velocity is the rate of change of distance).
• In our present case, $\rm{s \,=\, f(t)\,=\,\frac{1}{2}. 3. t^2 \,=\, 1.5 t^2}$
• So $\rm{f'(t) \,=\, \frac{ds}{dt} \,=\, \frac{d}{dt} \left(1.5 t^2 \right) \,=\, 1.5 \times 2t = 3t}$
6. Let us see an example:
The velocity when the stop-watch reading is 0.3 s, is (3 × 0.3) = 0.9 $\rm{m s^{-1}}$
• From the graph, we see that:
Rate of change (velocity in this case) is same as slope of the tangent.

---

Let us see some solved examples:

Solved example 22.1
If the radius of a circle is changing, what is the rate of change of the area of that circle, w.r.t radius?
Solution:
1. The area of circle is given by the equation: $\rm{A~=~\pi r^2}$
• Where 'A' is the area and 'r' is the radius.
• So it is clear that, if the radius changes, the area will also change.
2. Using the above equation, we can draw a graph which shows the variation of area. It is given in fig.22.2 below:

Fig.22.2

3. We see that, the graph is a curve. That means, the rate of change of area is not uniform. If the rate of change was uniform, we would have obtained a straight line graph.
• Due to the non-uniform nature, we can find only the instantaneous rate of change $\rm{\frac{dA}{dr}}$
• So $\rm{f'(r) \,=\, \frac{dA}{dr} \,=\, \frac{d}{dr} \left(\pi r^2 \right) \,=\, \pi \times 2r = 2 \pi r}$

4. Let us see an example:
The rate of change of area when the radius is 0.2 m is:
(2 × π × 0.2) = 1.26 $\rm{m^2 m^{-1}}$
• That means, for every one meter change of radius, the area changes by 1.26 m².
• From the graph, we see that:
Rate of change of area is same as slope of the tangent at (r=0.2).

---

• The instantaneous rate of change of y w.r.t x, at a particular point x₀, is denoted as: $\rm{\left. \frac{dy}{dx} \right|_{x =x_0}}$

• It can be also denoted as $\rm{f'(x_0)}$

• So in the above example, we can write:
$\rm{\left. \frac{dA}{dr} \right|_{r = 0.2}~=~1.26 ~ m^2 m^{-1}}$

---

Solved example 22.2
The radius of a circle is increasing at the rate of 4 cm s⁻¹. what is the rate of change of the area of that circle, w.r.t time, at the instant when the radius is 10 cm?
Solution:
1. The area of circle is given by the equation: $\rm{A~=~\pi r^2}$
• Where 'A' is the area and 'r' is the radius.
• So it is clear that, if the radius changes, the area will also change.
2. The radius is increasing continuously. So the area will also increase continuously.
• In the previous example, we saw that, the instantaneous rate of change of area w.r.t radius is $\rm{\frac{dA}{dr}}$.
• In this example, we are asked to find the instantaneous rate of change of area w.r.t time, which  is $\rm{\frac{dA}{dt}}$. 
3. So we need to bring 't' also into the calculations.
• The rate of increase of radius is given as 4 cm s⁻¹. We can write:
   ♦ When t = 0, r = 0
   ♦ When t = 1, r = 4 cm
   ♦ When t = 2, r = 8 cm
   ♦ When t = 3, r = 12 cm
so on . . .
• That means, at any time t, the radius will be 4t.

4. Now we can modify the equation for area:
$\rm{A~=~\pi r^2 ~=~ \pi (4t)^2}$
• This gives: $\rm{\frac{dA}{dt} ~=~16 \pi \times 2t ~=~32 \pi t}$
• Using this result, we can calculate the rate of increase of area w.r.t time, at any instant.

5. We are asked to calculate the rate of increase of area w.r.t time, at the instant when radius is 10 cm.
• So we first need to find that instant.
   ♦ We have: r = 4t = 10 cm.
   ♦ So t = 10/4 = 2.5 s
• That means, the radius will be 10 cm when t = 2.5 s.
• Substituting this t in (4), we get:
$\rm{\frac{dA}{dt} ~=~32 \pi t = 32 \times \pi \times 2.5 ~=~80 \pi ~cm^2 s^{-1}}$ 
• That is., $\rm{\left. \frac{dA}{dt} \right|_{r = 10}~=~80 \pi ~ cm^2 s^{-1}}$
• We can say that:
At the instant when radius is 10 cm, the area is increasing at the rate of $\rm{80 \pi ~ cm^2}$ per second.
At any other instant, the rate will be different. 

Solved example 22.3
The volume of a cube is increasing at the rate of 9 cm³ s⁻¹. what is the rate of change of the surface area of that cube, w.r.t time, at the instant when the length of an edge is 10 cm?
Solution:
1. Let V be the volume, S the surface area and l the length of edge.
• We can easily calculate $\rm{\frac{dS}{dl}}$ because, S = 6l².
• But we are asked to find $\rm{\frac{dS}{dt}}$ 

2. So we need to bring 't' also into the calculations.
• The rate of increase of volume is given as 9 cm³ s⁻¹. We can write:
   ♦ When t = 0, V = 0 cm³.
   ♦ When t = 1, V = 9 cm³.
   ♦ When t = 2, V = 18 cm³.
   ♦ When t = 3, V = 27 cm³.
so on . . .
• That means, at any time t, the volume will be 9t.
• That means, at any time t, the length of an edge l can be obtained using the equation: $\rm{l~=~(V)^{\frac{1}{3}}~=~(9t)^{\frac{1}{3}}}$.

3. So at any time t, the surface area S can be obtained using the equation:
$\rm{S ~=~6 \times l^2~=~6 \left[(9t)^{\frac{1}{3}} \right]^2~=~6 \left[9^{\frac{2}{3}} t^{\frac{2}{3}} \right] }$

4. Then the rate of change of S w.r.t to time can be obtained using the equation:
$\rm{\frac{dS}{dt}\,=\,6 \times 9^{\frac{2}{3}} \times \frac{2}{3} \times t^{\frac{-1}{3}}}$

• Using this result, we can calculate the rate of increase of surface area w.r.t time, at any instant.

5. We are asked to calculate the rate of increase of surface area w.r.t time, at the instant when length of edge is 10 cm.
• So we first need to find that instant.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l}    & {~=~}    &{(9t)^{\frac{1}{3}}}    \\
{~\color{magenta}    2    }    &{\implies}    &{10}    & {~=~}    &{(9t)^{\frac{1}{3}}}    \\
{~\color{magenta}    3    }    &{\implies}    &{10^3}    & {~=~}    &{9t}    \\
{~\color{magenta}    4    }    &{\implies}    &{t}    & {~=~}    &{\frac{1000}{9}}    \\
\end{array}$                           
• That means, the length of edge will be 10 cm when t = 1000/9 s.
• Substituting this t in (4), we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dS}{dt}}    & {~=~}    &{6 \times 9^{\frac{2}{3}} \times \frac{2}{3} \times t^{\frac{-1}{3}}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{6 \times 9^{\frac{2}{3}} \times \frac{2}{3} \times \left({\frac{1000}{9}} \right)^{\frac{-1}{3}}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{2 \times 9^{\frac{2}{3}} \times 2 \times \left({\frac{9}{1000}} \right)^{\frac{1}{3}}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{2 \times 9 \times 2 \times \left({\frac{1}{10}} \right)}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{36}{10}}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{3.6~ \rm{cm^2 \,s^{-1}}}    \\
\end{array}$

• That is., $\rm{\left. \frac{dS}{dt} \right|_{l = 10}~=~3.6 ~ cm^2 s^{-1}}$

• We can say that:
At the instant when length of edge is 10 cm, the surface area is increasing at the rate of $\rm{3.6 ~ cm^2}$ per second.
At any other instant, the rate will be different.  

Solved example 22.4
The length x of a rectangle is decreasing at the rate of 3 cm / minute and width y is increasing at the rate of 2 cm / minute. When x = 10 cm and y = 6 cm, find the rates of changes of (a) the perimeter and (b) area of the rectangle.
Solution:
Part (a):
1. Suppose that, the initial length is x₀.
• Given that, x is decreasing at the rate of 3 cm / minute. So we can write:
   ♦ When t = 1, x = x₀ − (3  × 1)
   ♦ When t = 2, x = x₀ − (3  × 2)
   ♦ When t = 3, x = x₀ − (3  × 3)
so on . . .
• Thus, the length at any time 't' is given by:
x = x₀ − (3  × t)

2. Suppose that, the initial width is y₀.
• Given that, y is increasing at the rate of 2 cm / minute. So we can write:
   ♦ When t = 1, y = y₀ + (2  × 1)
   ♦ When t = 2, y = y₀ + (2  × 2)
   ♦ When t = 3, y = y₀ + (2  × 3)
so on . . .
• Thus, the width at any time 't' is given by:
y = y₀ + (2  × t)

3. Perimeter at any time 't' will be given by:
P= 2[x+y] = 2[(x₀ − 3t) + (y₀ + 2t)] = 2[x₀ + y₀ − t] = 2(x₀ + y₀)− 2t

4. Rate of change of perimeter w.r.t time, at any instant, will be given by:
$\rm{\frac{dP}{dt}\,=\, \frac{d}{dt} \left[2(x_0 + y_0) - 2t \right]\,=\,-2}$
• We see that, the rate is a constant. That means, the rate is independent of x, y or t.
• We are asked to find the rate when x = 10 cm and y = 6 cm. At this length and breadth also, the rate will be −2 cm/minute.

• We can say that:
At the instant when x = 10 cm and y = 6 cm, the perimeter is decreasing at the rate of $\rm{2 ~ cm}$ per minute.
At any other instant, the rate will be same.  

Part (b):
1. Area at any time 't' will be given by:
$\rm{A\,=\, xy \,=\,(x_0 - 3t)(y_0 + 2t)\,=\,(x_0 y_0 + 2 x_0 t - 3 y_0 t - 6 t^2)}$
⇒ $\rm{A\,=\,x_0 y_0 + (2 x_0  - 3 y_0) t - 6 t^2}$

2. Rate of change of area w.r.t time, at any instant, will be given by:
$\rm{\frac{dA}{dt}\,=\, \frac{d}{dt} \left[x_0 y_0 + (2 x_0  - 3 y_0) t - 6 t^2 \right]\,=\,2 x_0  - 3 y_0 - 12 t}$

3. We are asked to find the dA/dt at the instant when x = 10 cm and y = 6 cm.
• So we first need to find the instant at which x becomes 10 cm and y becomes 6 cm.
• We can write:
(i) 10 = x₀ − 3t
(ii) 6 = y₀ + 2t
• The two equations can be rearranged as shown below:
(iii) → [(i) × 2] → 20 = 2x₀ − 6t 
(iv) → [(ii) × 3] → 18 = 3y₀ + 6t
(v) → [(iii) − (iv)] → 2 = 2x₀ − 3y₀ − 12t    
⇒ 2x₀ − 3y₀ = 2 + 12t

4. Substituting the above result in (2), we get:
$\rm{\frac{dA}{dt}\,=\,2 + 12t - 12 t \,=\,2}$

• We can say that:
At the instant when x = 10 cm and y = 6 cm, the area is increasing at the rate of $\rm{2 ~ cm^2}$ per minute.
At any other instant, the rate will be different.

Solved example 22.5
The total cost C(x) in Rupees, associated with the production of x units of an item is given by:
C(x) = 0.005 x³ − 0.02x² + 30x + 5000
Find the marginal cost when 3 units are produced. Marginal cost is the instantaneous rate of change of total cost at any level of output.
Solution:
Let us write some basic details about marginal cost. It can be written in 4 steps:
1. Consider the production of an item. By 'item', we mean products like LED bulb, bicycle, microwave oven etc.,

2. Take any one particular item. To produce 'x' no. of that item, the factory has to spend a certain amount of money.
• This amount (in Rupees) is given by:
C(x) = 0.005 x³ − 0.02x² + 30x + 5000

3. We see that, there are four terms. Why is there four terms?
• One term may be related to the cost of raw materials required for production.
• Another term may be related to the wages of workers involved in the production.
For example, if Rupees 30/- is the labor charge for one number, then 30x has to be spend for the production of x numbers. So this term is related to the labor department.
• Another term may be related to the cost of electricity, water supply etc.,

4. However, we need not worry about what each term represent. We are asked to find the marginal cost.
• Marginal cost is the instantaneous rate of change of cost w.r.t number of items produced.
• So marginal cost is $\rm{\frac{dC}{dx}}$.
• In the numerator, we have change of cost. In the denominator, we have change in number. But we need not calculate numerator and denominator separately. We know how to calculate $\rm{\frac{dC}{dx}}$ directly.

---

Now we can do the calculations:
• We have:
$\rm{\frac{dC}{dx}\,=\,\frac{d}{dx} \left[0.005 x^3 - 0.02 x^2 + 30 x + 5000 \right] \,=\, 0.015 x^2 - 0.04 x + 30}$
• So when 3 units are produced, we get:
$\rm{\left. \frac{dC}{dx} \right|_{x = 3}\,=\,0.015 (3)^2 - 0.04 (3) + 30 \,=\,0.135 - 0.12 + 30 = \text{Rs} \, 30.015 \, \text{per number}}$

• We can say that:
At the instant when number of items produced is 3, the marginal cost is Rs 30.015.
At any other instant, the marginal cost will be different.   

Solved example 22.6
The total revenue R(x) in Rupees, received from the sale of x units of a product is given by:
R(x) = 3 x² + 36x + 5
Find the marginal revenue when 5 units are sold. Marginal revenue is the instantaneous rate of change of total revenue at any level of sale.
Solution:
Let us write some basic details about marginal revenue. It can be written in steps:
1. Consider the sale of a product. By 'product', we mean products like LED bulb, bicycle, microwave oven etc.,

2. Take any one particular product. When 'x' no. of that product is sold, the factory gets a certain amount of money.
• This amount (in Rupees) is given by:
R(x) = 3 x² + 36x + 5

3. We see that, there are three terms. Why is there three terms?
• One term may be the actual amount received from the customer.
• Another term may be related to the amount received as commission.

4. However, we need not worry about what each term represent. We are asked to find the marginal revenue.
• Marginal revenue is the instantaneous rate of change of revenue w.r.t number of items sold.
• So marginal revenue is $\rm{\frac{dR}{dx}}$.
• In the numerator, we have change of revenue. In the denominator, we have change in number. But we need not calculate numerator and denominator separately. We know how to calculate $\rm{\frac{dR}{dx}}$ directly.

---

Now we can do the calculations:
• We have:
$\rm{\frac{dR}{dx}\,=\,\frac{d}{dx} \left[3 x^2 + 36 x + 5 \right] \,=\, 6 x + 36}$
• So when 5 units are sold, we get:
$\rm{\left. \frac{dR}{dx} \right|_{x = 5}\,=\,6(5) + 36 \,=\, = \text{Rs} \, 66 \, \text{per number}}$

• We can say that:
At the instant when number of items sold is 5, the marginal revenue is Rs 66.
At any other instant, the marginal revenue will be different.   

---

The link below gives a few more solved examples:

Exercise 22.1

---

In the next section, we will see increasing and decreasing functions.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

21.22 - More Miscellaneous Examples

In the previous section, we saw some miscellaneous examples. In this section, we will see a few more examples.

Solved example 21.68
Find f'(x) if f(x) = $\rm{(\sin x)^{\sin x}}$ for all 0<x<π.
Solution:

 

Solved example 21.69
For a positive constant a, find $\rm{\frac{dy}{dx}}$ where:
$\rm{y = a^{t + 1/t}~\text{and}~x = \left(t + 1/t \right)^a}$
Solution:
1. First we will find dy/dt:

2. Next we will find dx/dt

3. Now we can find dy/dx

 

Solved example 21.70
Differentiate $\rm{\sin^2 x~\text{w.r.t}~e^{\cos x}}$.
Solution:
1. Let $\rm{y = \sin^2 x~\text{and}~x = e^{\cos x}}$
2. We want the derivative of
$\rm{\sin^2 x~\text{w.r.t}~e^{\cos x}}$
• That is., we want the derivative of
$\rm{\sin^2 x~\text{w.r.t}~u}$
• That is., we want the derivative of
$\rm{y~\text{w.r.t}~u}$
• That is., we want $\rm{\frac{dy}{du}}$
3. First we will find dy/dx:
$\rm{y = \sin^2 x}$
So we have:$\rm{\frac{dy}{dx}~=~2 \sin x \cos x}$

4. Next we will find du/dx:

 

5. Now we can find dy/du:

---

The link below gives a few more miscellaneous examples:

Miscellaneous Exercise

---

In the next chapter, we will see application of derivatives.

Previous

Contents


Next

Copyright©2024 Higher secondary mathematics.blogspot.com

21.21 Miscellaneous Examples

In the previous section, we completed a discussion on Rolle's theorem and mean value theorem. In this section, we will see some miscellaneous examples.

Solved example 21.66
Differentiate w.r.t x the following functions:
(i) $\rm{\sqrt{3x+2}~+~\frac{1}{\sqrt{2x^2 + 4}}}$
(ii) $\rm{e^{\sec^2 x}~+~3 \cos^{-1} x}$
(iii) $\rm{\log_7 (\log x)}$
Solution:
Part (i):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\sqrt{3x+2}~+~\frac{1}{\sqrt{2x^2 + 4}}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left(3x+2 \right)^{1/2} ~+~\left(2x^2 + 4 \right)^{-1/2}} \\
{~\color{magenta} 3 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{1/2 \left(3x+2 \right)^{-1/2}.3 ~+~(-1/2)\left(2x^2 + 4 \right)^{-3/2}.4x} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{3}{2 \sqrt{3x+2}}~-~\frac{2x}{\left(\sqrt{2x^2 + 4} \right)^3}} \\
\end{array}$

Part (ii):
1. First we will split the given function
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{e^{\sec^2 x}~+~3 \cos^{-1} x} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{u+v} \\
{~\color{magenta} 3 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{du}{dx}~+~\frac{dv}{dx}} \\
\end{array}$

2. Now we will find $\rm{\frac{du}{dx}}$:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{u} & {~=~} &{e^{\sec^2 x}} \\
{~\color{magenta} 2 } &{\implies} &{\log u} & {~=~} &{\sec^2 x.\log_e e} \\
{~\color{magenta} 3 } &{\implies} &{\log u} & {~=~} &{\sec^2 x} \\
{~\color{magenta} 4 } &{\implies} &{\frac{d}{dx} \left[\log u \right]} & {~=~} &{\frac{d}{dx} \left[\sec^2 x \right]} \\
{~\color{magenta} 5 } &{\implies} &{\frac{1}{u}.\frac{du}{dx}} & {~=~} &{2 \sec x . \sec x \tan x} \\
{~\color{magenta} 6 } &{\implies} &{\frac{du}{dx}} & {~=~} &{u. 2 \sec x . \sec x \tan x} \\
{~\color{magenta} 7 } &{{}} &{{}} & {~=~} &{2 e^{\sec^2 x} \sec^2 x \tan x} \\
\end{array}$

3. Now we will find $\rm{\frac{dv}{dx}}$:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{v} & {~=~} &{3 \cos^{-1} x} \\
{~\color{magenta} 2 } &{\implies} &{\frac{dv}{dx}} & {~=~} &{3 \frac{d}{dx} \left[\cos^{-1} x \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{3. \frac{(-1)}{\sqrt{1 – x^2}}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{-3}{\sqrt{1 – x^2}}} \\
\end{array}$

4. So based on (1), we get:
$\rm{\frac{dy}{dx}~=~\frac{du}{dx}~+~\frac{dv}{dx}}$

= $\rm{2 e^{\sec^2 x} \sec^2 x \tan x~-~\frac{3}{\sqrt{1 – x^2}}}$

Part (iii):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\log_7 (\log x)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{\log_e (\log x)}{\log_e 7}} \\
{~\color{magenta} 3 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{1}{\log_e 7} \frac{d}{dx} \left[\log_e (\log x) \right]} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{1}{\log_e 7} .\frac{1}{\log x}.\frac{1}{x}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{1}{x \log 7 \log x}} \\
\end{array}$


◼ Remarks:
• 2 (magenta color): Here we apply change of base rule.
• 5 (magenta color): When the base of logarithm is not specified, it implies that, the base is e.

Solved example 21.67
Differentiate w.r.t x the following functions:
(i) $\rm{\cos^{-1} (\sin x)}$
(ii) $\rm{\tan^{-1} \left(\frac{\sin x}{1 + \cos x} \right)}$
(iii) $\rm{\sin^{-1} \left(\frac{2^{x+1}}{1 + 4^x} \right)}$
Solution:
Part (i):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\cos^{-1} (\sin x)} \\
{~\color{magenta} 2 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{-1}{\sqrt{1 - \sin^2 x}}.\frac{d}{dx} (\sin x)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{-1}{\sqrt{\cos^2 x}}.\cos x} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{-1} \\
\end{array}$

Alternate method:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\cos^{-1} (\sin x)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\cos^{-1} \left[\cos\left(\frac{\pi}{2} ~-~x \right) \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{\pi}{2} ~-~x} \\
{~\color{magenta} 4 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{-1} \\
\end{array}$

Part (ii): $\rm{\tan^{-1} \left(\frac{\sin x}{1 + \cos x} \right)}$
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\tan^{-1} \left(\frac{\sin x}{1 + \cos x} \right)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\tan^{-1} \left(\tan \frac{x}{2} \right)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{x}{2}} \\
{~\color{magenta} 4 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{1}{2}} \\
\end{array}$


◼ Remarks:
• 2 (magenta color): Here we use identities:
$\rm{\cos x ~=~ 1 - 2 \sin^2 \frac{x}{2}}$
$\rm{\cos x ~=~ 2 \cos^2 \frac{x}{2} - 1}$

Combining these two identities, we get:
$\rm{\tan \frac{x}{2}~=~\frac{\sin x}{1 + \cos x}}$

Part (iii): $\rm{\sin^{-1} \left(\frac{2^{x+1}}{1 + 4^x} \right)}$

1. First we will simplify the expression.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Let}~2^x}    & {~=~}    &{t}    \\
{~\color{magenta}    2    }    &{{}}    &{\text{Given:}~\frac{2^{x+1}}{1+4^x}}    & {~=~}    &{\frac{2.2^x}{1+(2.2)^x}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{2.2^x}{1+2^x .2^x}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{2t}{1+t^2}}    \\
\end{array}$

2. Next we will write 't' in terms of 𝜃.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Let}~t}    & {~=~}    &{\tan \frac{\theta}{2}}    \\
{~\color{magenta}    2    }    &{{}}    &{\text{Then:}~\frac{2t}{1+t^2}}    & {~=~}    &{\sin \theta}    \\
\end{array}$

(Identity 15. List of identities can be seen here) 

3. Next we will write 'y' in terms of 𝜃.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\sin^{-1} \left(\frac{2^{x+1}}{1 + 4^x} \right)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\sin^{-1} \left(\frac{2t}{1 + t^2} \right)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\sin^{-1}(\sin \theta)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\theta}    \\
\end{array}$

4. Next we will write 𝜃 in terms of 'x'.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2^x}    & {~=~}    &{t}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\tan \frac{\theta}{2}}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{\theta}{2}}    & {~=~}    &{\tan^{-1}(2^x)}    \\
{~\color{magenta}    4    }    &{\implies}    &{\theta}    & {~=~}    &{2. \tan^{-1}(2^x)}    \\
\end{array}$

5. Based on (3), we can now connect 'y' and 'x'.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{2. tan^{-1}(2^x)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{2. \frac{1}{1 + (2^x)^2}.\frac{d}{dx}(2^x)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{2. \frac{1}{1 + (2^2)^x}.\frac{d}{dx}(2^x)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{2}{1 + 4^x}.\frac{d}{dx}(2^x)}    \\
\end{array}$

6. So our next task is to find the derivative of $\rm{2^x}$.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{u}    & {~=~}    &{2^x}    \\
{~\color{magenta}    2    }    &{\implies}    &{\log u}    & {~=~}    &{x \log 2}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{1}{u}.\frac{du}{dx}}    & {~=~}    &{1. \log 2}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{du}{dx}}    & {~=~}    &{u.\log 2}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{2^x \log 2}    \\
\end{array}$

7. So from (5), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{2}{1 + 4^x}.\frac{d}{dx}(2^x)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{2}{1 + 4^x}.(2^x) \log 2}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{2^{x+1} \log 2}{1 + 4^x}}    \\
\end{array}$

---

In the next section, we will see a few more solved examples.

Previous

Contents


Next

Copyright©2024 Higher secondary mathematics.blogspot.com

21.20 - Mean Value Theorem

In the previous section, we completed a discussion on second order differentiation. In this section, we will see Rolle's theorem and  mean value theorem.

The basic details about Rolle's theorem can be written in 4 steps:

1. In the fig.21.23 below, the red curve is the graph of a function f.

Fig.21.23

2. The function f satisfies the following four conditions:
(i) f is a real valued function.
(ii) f is continuous in the interval [a,b]
(iii) f is differentiable in the interval (a,b)
(iv) f(a) = f(b)
• Note that, the solid green line which passes through both (a, f(a)) and (b, f(b)) is horizontal.
• Since the solid green line is horizontal, f(a) = f(b)
3. If the four conditions are satisfied, we can write:
There is atleast one point c in the interval (a,b), such that f'(c) = 0
• Note that, a tangent (yellow line) is drawn to the red curve at (c, f(c)). This tangent is horizontal. Since this tangent is horizontal, we can say that, slope of the tangent at c is zero. In other words, derivative at c is zero.
4. Another example is shown in fig.21.24 below:

Fig.21.24

• We see that:
(i) g is a real valued function.
(ii) g is continuous in the interval [a,b]
(iii) g is differentiable in the interval (a,b)
(iv) g(a) = g(b)
• In this case, there are two points c and d, where the derivative is zero.

---

Solved example 21.64
Verify Rolle's theorem for the function f(x) = x² + 2
when a = −2 and b = 2
Solution:
1. Let us check whether the four conditions are satisfied:
(i) Given f is a real valued function.
(ii) f is continuous in the interval [−2, 2]
(iii) f is differentiable in the interval (−2, 2)
(iv) f(−2) = (−2)² + 2 = 6 = f(2) = (2)² + 2
• So the four conditions are satisfied.
2. Since the four conditions are satisfied, Rolle's theorem is applicable.
• That means, there exists atleast one number c, in the interval (−2, 2) such that, f'(c) = 0. Our aim is to find this c.
3. We have: f'(x) = 2x
• Equating f'(x) to zero, we get: 2x = 0
⇒ x = 0
4. That means, f'(0) = 0.
Therefore, c = 0
• c = 0 indeed lies in the interval (−2, 2)
5. The graph is shown in fig.21.25 below:

Fig.21.25

 

• The tangent at c = 0 is drawn in yellow color.
• This tangent is horizontal, which means, slope of the tangent is zero. In other words, the derivative at c = 0, is zero.

---

Now we will see mean value theorem (M.V.T). The basic details can be written in 5 steps:

1. In the fig.21.26 below, the red curve is the graph of a function f.

Fig.21.26

2. The function f satisfies the following three conditions:
(i) f is a real valued function.
(ii) f is continuous in the interval [a,b]
(iii) f is differentiable in the interval (a,b)
3. If the three conditions are satisfied, we can write:
There is atleast one point c in the interval (a,b), such that:
$\rm{f'(c)~=~\frac{f(b) - f(a)}{b-a}}$
4. Let us see the significance of the quantity $\rm{\frac{f(b) - f(a)}{b-a}}$.
It can be written in 5 steps:
(i) The vertical distance between (a, f(a)) and (b, f(b)) is [f(b) − f(a)]
(ii) The horizontal distance between (a, f(a)) and (b, f(b)) is [b−a]
(iii) So $\rm{\frac{f(b) - f(a)}{b-a}}$ is the slope of the line joining (a, f(a)) and (b, f(b))
(iv) But the line joining (a, f(a)) and (b, f(b)), is the secant between (a, f(a)) and (b, f(b)).
(v) So we can write:
$\rm{f'(c)~=~\frac{f(b) - f(a)}{b-a}}$ is the slope of the secant between (a, f(a)) and (b, f(b)).
5. So based on (3), we can write:
If the three conditions are satisfied,
There is atleast one point c in the interval (a,b), such that:
the derivative at c is same as the slope of the secant between (a, f(a)) and (b, f(b)).

---

Solved example 21.65
Verify Mean Value Theorem for the function f(x) = x²
when a = 2 and b = 4
Solution:
1. Let us check whether the three conditions are satisfied:
(i) Given f is a real valued function.
(ii) f is continuous in the interval [2, 4]
(iii) f is differentiable in the interval (2, 4)
• So the three conditions are satisfied.
2. Since the three conditions are satisfied, M.V.T is applicable.
• That means, there exists atleast one number c, in the interval (2, 4) such that, $\rm{f'(c)~=~\frac{f(4) - f(2)}{4-2}}$. Our aim is to find this c.
3. The R.H.S can be calculated as shown below:
$\rm{\frac{f(4) - f(2)}{4-2}~=~\frac{4^2 - 2^2}{4-2}~=~\frac{16 - 4}{4-2}~=~\frac{12}{2}~=~6}$
4. We have: f'(x) = 2x
• Equating f'(x) to 6, we get: 2x = 6
⇒ x = 3
5. That means, f'(3) = 6.
Therefore, c = 3
• c = 3 indeed lies in the interval (2, 4)
6. The graph (red curve) is shown in fig.21.27 below:

Fig.21.27

• The scale of x and y axes are adjusted so as to make the curvature of the graph visible.
• The secant (yellow line) is drawn between a = 2 and b = 4
• The tangent (green line) is drawn at c = 3
• The tangent is parallel to the secant. Both have the slope 6.

---

The link below gives a few more solved examples:

Exercise 21.8

---

In the next section, we will see some miscellaneous examples.

Previous

Contents


Next

Copyright©2024 Higher secondary mathematics.blogspot.com