22.14 - Analytical Method For Finding Absolute Maximum and Absolute Minimum

In the previous section, we we completed a discussion on critical points. In this section, we will see the analytical method to find absolute maximum and absolute minimum.

• We know how to find all critical points, with out the help of a graph.
Let x₁, x₂, x₃, . . .  be the critical points.
• The next step naturally, is to find the f values at those points.
So we get: f(x₁), f(x₂), f(x₃), . . .
• Then we compare those f values. After the comparison, we are inclined to believe that:
   ♦ The largest f value is the absolute maximum.
   ♦ The smallest f value is the absolute minimum.
• But before we make such a conclusion, there is another aspect which should be examined. It can be written in 5 steps:

1. Fig.22.38 below shows the graph of a function f.

Fig.22.38

2. x₁ and x₂ are the two critical points.
3. From the graph, it is clear that, f(x₁) will be larger than f(x₂). So we are inclined to conclude that, x₁ is the absolute maximum and x₂ is the absolute minimum.
4. But what if we are to find the extrema in the interval [a,b]?
In the fig., f(a) and f(b) are clearly marked.
• We see that:
   ♦ f(b) is larger than f(x₁).
   ♦ So the absolute maximum is f(b). Not f(x₁).
• We see that:
   ♦ f(a) is smaller than f(x₂).
   ♦ So the absolute minimum is f(a). Not f(x₂)
5. While finding the critical points, the end points a and b will not show up because, the derivatives at those points are not zero. Also, derivatives exist at those points.
• So it is clear that, before finalizing absolute maximum and absolute minimum, we need to check the endpoints also.

Now we will see some solved examples

Solved Example 22.49
Find the absolute maximum value and absolute minimum value of the following function in the given intervals:
(i) f(x) = x³, x ∈ [−2,2]
(ii) f(x) = sin x + cos x, x ∈ [0,π]
(iii) f(x) = 4x − (1/2)x², x ∈ [−2,9/2]
(iv) f(x) = (x−1)² + 3, x ∈ [−3,1]
Solution:
Part (i): f(x) = x³, x ∈ [−2,2]
Step I: Finding the critical points and evaluating f
1. We have: $\rm{f'(x)\,=\,3x^2}$
2. Equating f'(x) to zero, we get:
3x² = 0
⇒ x² = 0
⇒ x = 0
3. So the point in category I is: x = 0
4. We obtained f'(x) = 3x²
• This function is a polynomial function. It is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 0
6. Evaluating f at the critical point, we get:
f(0) = (0)³ = 0

Step II: Evaluating f at end points
1. f(−2) = (−2)³ = −8
2. f(2) = (2)³ = 8

Step III: Comparing the f values
1. Absolute maximum is 8, which occurs at x = 2
2. Absolute minimum is −8, which occurs at x = −2

• Fig.22.39 below shows the graph:

Fig.22.39

• The dashed magenta vertical lines represent the end points.
• We obtained the extrema because we checked the end points also. If rely solely on the critical points, we will not get the actual extrema.

Part (ii): f(x) = sin x + cos x, x ∈ [0,π]
Step I: Finding the critical points and evaluating f
1. We have: $\rm{f'(x)\,=\,\cos x - \sin x}$
2. This derivative must be equated to zero. For that, it should be rearranged as follows:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(x)}    & {~=~}    &{\cos x - \sin x}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\sin (\pi/2 -x) - \sin x}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{2 \cos \left(\frac{\pi/2 - x + x}{2} \right) \sin \left(\frac{\pi/2 - x - x}{2} \right)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{2 \cos \left(\frac{\pi}{4} \right) \sin \left(\frac{\pi}{4} - x \right)}    \\
\end{array}$                           

• Equating f'(x) to zero, we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2 \cos \left(\frac{\pi}{4} \right) \sin \left(\frac{\pi}{4} - x \right)}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{\sin \left(\frac{\pi}{4} - x \right)}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{\pi}{4} - x}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{x}    & {~=~}    &{\frac{\pi}{4}}    \\
\end{array}$

◼ Remarks:
(4) Magenta color: Here we do not require the general solution because, x lies within the given interval [0,π]

3. So the point in category I is: x = π/4
4. We obtained f'(x) = cos x − sin x
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = π/4
6. Evaluating f at the critical point, we get:
f(π/4) = sin(π/4) + cos(π/4)
= $\rm{\frac{1}{\sqrt2}+ \frac{1}{\sqrt2} ~=~\frac{2}{\sqrt2}~=~\sqrt2}$

Step II: Evaluating f at end points
1. f(0) = sin 0 + cos 0 = (0+1) = 1
2. f(π) = sin π + cos π (0−1) = −1

Step III: Comparing the f values
1. Absolute maximum is √2, which occurs at x = π/4
2. Absolute minimum is −1, which occurs at x = π

• Fig.22.40 below shows the graph:

Fig.22.40

• The y-axis and the dashed magenta vertical line represent the end points.
• Absolute maximum occurs at the critical point.
• Absolute minimum occurs at the right end point.

Part (iii): f(x) = 4x − (1/2)x², x ∈ [−2,9/2]
Step I: Finding the critical points and evaluating f
1. We have: $\rm{f'(x)\,=\,4 - x}$
2. This derivative must be equated to zero. We get:
4 − x = 0
⇒ x = 4

3. So the point in category I is: x = 4
4. We obtained f'(x) = 4 − x
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 4
6. Evaluating f at the critical point, we get:
f(4) = 4(4) − (1/2) (4)² = (16 − 8) = 8

Step II: Evaluating f at end points
1. f(−2) = 4(−2) − (1/2) (−2)² = (−8 − 2) = −10
2. f(9/2) = 4(9/2) − (1/2) (9/2)² = (18 − 81/8) =$\rm{7 \frac{7}{8}}$

Step III: Comparing the f values
1. Absolute maximum is 8, which occurs at x = 4
2. Absolute minimum is −10, which occurs at x = −2

• Fig.22.41 below shows the graph:

Fig.22.41

• The dashed magenta vertical lines represent the end points.
• Absolute maximum occurs at the critical point.
• Absolute minimum occurs at the left end point.

Part (iv): f(x) = (x−1)² + 3, x ∈ [−3,1]
Step I: Finding the critical points and evaluating f
1. We have: $\rm{f'(x)\,=\,2(x-1)\,=\,2x - 2}$
2. This derivative must be equated to zero. We get:
2x − 2 = 0
⇒ 2x = 2
⇒ x = 1

3. So the point in category I is: x = 1
4. We obtained f'(x) = 2x − 2
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 1
6. Evaluating f at the critical point, we get:
f(1) = (1−1)² + 3 = (0+3) = 3

Step II: Evaluating f at end points
1. f(−3) = (−3 −1)² + 3 = (16+3) = 19
2. f(1) = (1 −1)² + 3 = (0+3) = 3

Step III: Comparing the f values
1. Absolute maximum is 19, which occurs at x = −3
2. Absolute minimum is 3, which occurs at x = 1

• Fig.22.42 below shows the graph:

Fig.22.42

• The dashed magenta vertical lines represent the end points.
• Absolute minimum occurs at the critical point.
• Critical point is same as the right end point.
• Absolute maximum occurs at the left end point.

Solved Example 22.50
Find the absolute maximum value and absolute minimum value of the function:
$\rm{f(x)\,=\,\sqrt{x}\,-\,\sqrt{x^3}}$, x ∈ [0,4]
Solution:
Step I: Finding the critical points and evaluating f
1. First we find the derivative:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(x)}    & {~=~}    &{\sqrt{x}\,-\,\sqrt{x^3}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{x^{1/2}\,-\,x^{3/2}}    \\
{~\color{magenta}    3    }    &{\implies}    &{f'(x)}    & {~=~}    &{(1/2)x^{-1/2}\,-\,(3/2)x^{1/2}}    \\
\end{array}$                           

2. This derivative must be equated to zero.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(1/2)x^{-1/2}\,-\,(3/2)x^{1/2}}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{1}{2 \sqrt x} \,-\, \frac{3 \sqrt x}{2}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{1}{2} \,-\, \frac{3 x}{2}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{1 - 3x}{2}}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{1 - 3x}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{\implies}    &{3x}    & {~=~}    &{1}    \\
{~\color{magenta}    7    }    &{\implies}    &{x}    & {~=~}    &{\frac{1}{3}}    \\
\end{array}$                           

◼ Remarks:
(3) Magenta color: Here we multiply the whole equation by √x.

3. So the point in category I is: x = 1/3
4. We obtained $\rm{f'(x) \,=\, (1/2)x^{-1/2}\,-\,(3/2)x^{1/2}}$
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 1/3 = 0.3333
6. Evaluating f at the critical point, we get:
$\rm{f(1/3)\,=\,\sqrt{1/3}\,-\,\sqrt{(1/3)^3}~=~0.3849}$

Step II: Evaluating f at end points
1. $\rm{f(0)\,=\,\sqrt{0}\,-\,\sqrt{(0)^3}~=~0}$
2. $\rm{f(4)\,=\,\sqrt{4}\,-\,\sqrt{(4)^3}~=~-6}$

Step III: Comparing the f values
1. Absolute maximum is 0.3849, which occurs at x = 1/3
2. Absolute minimum is −6, which occurs at x = 4

• Fig.22.43 below shows the graph:

Fig.22.43

• The y-axis and the dashed magenta vertical line represent the end points.
• Absolute maximum occurs at the critical point.
• Absolute minimum occurs at the right end point.

---

In the next section, we will see shape of graph.

Previous

Contents


Copyright©2024 Higher secondary mathematics.blogspot.com

22.13 - Solved Examples on Critical Points

In the previous section, we saw critical points. We saw a solved example also. In this section, we will see a few more solved examples on critical points.

Solved Example 22.43
Find all critical points for the function:
$\rm{f(x)\,=\,2x^3 - 6x^2 + 6x + 5}$
Solution:
1. We have: $\rm{f'(x)\,=\,6x^2 - 12x + 6}$
2. Equating f'(x) to zero, we get:
6(x² − 2x + 1) = 0
⇒ 6(x² − 2x + 1) = 0
⇒ (x² − 2x + 1) = 0
⇒ (x − 1)² = 0
⇒ (x − 1) = 0
⇒ x = +1
3. So the point in category I is: x = 1
4. We obtained f'(x) = 6(x² − 2x + 1)
• This function is a polynomial function. It is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 1
6. Fig.22.32 shows the graph:

Fig.22.32

• From the graph, it is clear that, x=1 is a critical point but it is not an extremum.

Solved Example 22.44
Find all critical points for the function:
$\rm{f(x)\,=\,(2-8x)^4 (x^2 - 9)^3}$
Solution:
1. First we will write the derivative:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{(2-8x)^4 (x^2 - 9)^3}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(2-8x)^4 \left[3(x^2 - 9)^2 (2x)\right]~+~\left[4(2-8x)^3 (-8) \right] (x^2 - 9)^3}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(2-8x)^4 \left[(x^2 - 9)^2 (6x)\right]~-~\left[(32)(2-8x)^3 \right] (x^2 - 9)^3}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(2-8x)^3 (x^2 - 9)^2 \left[  (2-8x)  (6x)~-~(32)(x^2 - 9) \right]}    \\
{~\color{magenta}    5    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(2-8x)^3 (x^2 - 9)^2 \left[ 12x - 48 x^2 -32 x^2 + (32)(9) \right]}    \\
{~\color{magenta}    6    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(2-8x)^3 (x^2 - 9)^2 \left[ 12x - 80 x^2  + (32)(9) \right]}    \\
\end{array}$                           

2. Equating the derivative to zero, we get three equations:
(i) (2 − 8x)³ = 0
(ii) (x² − 9)² = 0
(iii) 12x − 80x² + (32)(9) = 0

• Solving the first equation, we get:
$\rm{x = \color{yellow} {\frac{1}{4}}}$
• Solving the second equation, we get:
$\rm{x = \color{yellow} {\pm 3}}$
• The third equation can be rearranged as:
3x − 20x² + 72 = 0
Solving this, we get:
$\rm{x = \color{yellow} {\frac{3 \pm \sqrt{5769}}{40}}}$
3. So there are five points in category I. They are written in yellow color.
4. Consider the derivative written in (1)
• This function is a polynomial function. It is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only five critical points are the ones we wrote in yellow color in (2)
6. Fig.22.33 shows the graph:

Fig.22.33

• From the graph, it is clear that:
   ♦ Points A, B and C are critical points but they are not extrema.
   ♦ Point D is a local minimum.
   ♦ Point E is the absolute minimum.
   ♦ This function does not have absolute maximum.

Solved Example 22.45
Find all critical points for the function:
$\rm{f(x)\,=\,4 \sqrt{x}\,-\, x^2}$
Solution:
1. First we will write the derivative:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{4 \sqrt{x}\,-\, x^2}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{4 (1/2)(x)^{-1/2}\,-\, 2x}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{2(x)^{-1/2}\,-\, 2x}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{2 \left[(x)^{-1/2}\,-\, x \right]}    \\
\end{array}$                           

2. Equating the derivative to zero, we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2 \left[(x)^{-1/2}\,-\, x \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{(x)^{-1/2}\,-\, x}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{\implies}    &{(x)^{-1/2}}    & {~=~}    &{x}    \\
{~\color{magenta}    4    }    &{\implies}    &{(x)^{-1}}    & {~=~}    &{x^2}    \\
{~\color{magenta}    5    }    &{\implies}    &{x^3}    & {~=~}    &{1}    \\
{~\color{magenta}    6    }    &{\implies}    &{x}    & {~=~}    &{1}    \\
\end{array}$

3. So the point in category I is: x = 1
4. We obtained $\rm{f'(x) = 2 \left[(x)^{-1/2}\,-\, x \right]}$
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 1
6. Fig.22.34 shows the graph:

Fig.22.34

• From the graph, it is clear that:
   ♦ x=1 is a critical point.
   ♦ Also, x=1 is an extremum. It is the absolute maximum.

Solved Example 22.46
Find all critical points for the function:
$\rm{f(x)\,=\,\frac{1}{x-1}}$
Solution:
1. First we will write the derivative:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\frac{1}{x-1}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{(x-1)^{-1}}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{(-1)(x-1)^{-2}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{(x-1)^2}}    \\
\end{array}$                           

2. Equating the derivative to zero, we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{-1}{(x-1)^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{\implies}    &{-1}    & {~=~}    &{0}    \\
\end{array}$                           
• This equation does not have any solution.

3. So there are no points in category I.
4. We obtained $\rm{f'(x) = \frac{-1}{(x-1)^2}}$
• This function is not defined at x = 1. But the f(x) is also not defined at x = 1. It is meaningless to say that f'(x) is not available at a point where f(x) itself is not available. So we need not consider the point x=1.
• Therefore, there will be no points in category II.
5. So there are no critical points for the given function
6. Fig.22.35 shows the graph:

Fig.22.35

• From the graph, it is clear that:
   ♦ When x decreases, f(x) approaches zero.
   ♦ When x increases, then also f(x) approaches zero.
   ♦ When x approaches 1 from the left, f(x) approaches −∞.
         ✰ So there is no absolute minimum.
   ♦ When x approaches 1 from the right, f(x) approaches +∞.
         ✰ So there is no absolute maximum.

Solved Example 22.47
Find all critical points for the function:
$\rm{f(x)\,=\,\tan x}$
Solution:
1. First we will write the derivative:
f'(x) = sec²x

2. Equating the derivative to zero, we get:
sec²x = 0                  
• This equation does not have any solution.

3. So there are no points in category I.
4. We obtained f'(x) = sec²x
• This function is not defined at x = π/2. But the f(x) is also not defined at x = π/2. It is meaningless to say that f'(x) is not available at a point where f(x) itself is not available. So we need not consider the point x = π/2.
• Therefore, there will be no points in category II.
5. So there are no critical points for the given function
6. Fig.22.36 shows the graph:

Fig.22.36

• From the graph, it is clear that:
   ♦ There is no absolute maximum.
   ♦ There is no absolute minimum.
   ♦ No tangent can be drawn in the horizontal direction.
         ✰ That means, there is no point where the derivative is zero.
   ♦ The function does not exist at $\rm{x = \frac{n \pi}{2}}$
         ✰ Where n is any integer.

Solved Example 22.48
Find all critical points for the function:
$\rm{f(x)\,=\,\sin^2 x}$
Solution:
1. First we will write the derivative:
f'(x) = 2 sin x cos x

2. Equating the derivative to zero, we get two equations:
(i) sin x = 0                  
(ii) cos x = 0
• Solving the first equation, we get:
x = 0, π, 2π, 3π, . . . so on
• Solving the first equation, we get:
x = π/2, 3π/2, 5π/2, . . . so on
• The above two results can be combined as:
$\rm{x \,=\,\frac{n \pi}{2}}$
   ♦ Where n is any integer.

3. So there are infinite points in category I.
4. We obtained f'(x) = 2 sin x cos x
• This function is defined for all real numbers. That means, there is no input x at which f'(x) is not defined.
• Therefore, there will be no points in category II.

5. So the critical points for the given function are given by:
$\rm{x \,=\,\frac{n \pi}{2}}$
   ♦ Where n is an integer.
6. Fig.22.37 shows the graph:

Fig.22.37

• From the graph, it is clear that:
   ♦ Absolute maximum occur at x = π/2, 3π/2, 5π/2, . . . so on.
   ♦ Absolute minimum occur at x = 0, 2π/2, 4π/2, 6π/2, . . . so on.

---

In the next section, we will see the analytical method for calculating Maxima and Minima.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com