In the previous section, we saw the derivative test which help us to find whether a given function is increasing or decreasing. We saw a solved example also. In this section, we will see a few more solved examples.
Solved example 22.9
Prove that the function given by f(x) = cos x is
(a) strictly decreasing in (0, π).
(b) strictly increasing in (π, 2π).
(c) neither increasing nor decreasing in (0,2π).
Solution:
First we write the derivative: f'(x) = −sin x
Part (a):
1. The interval that we need to consider in part (a) is (0,π).
2. This interval indicates I and II quadrants. In I and II quadrants, sine is always +ve. So −sin x will be −ve.
3. We can write:
f'(x) < 0. So f(x) is strictly decreasing in this interval.
4. Note that (0,π) is an open interval. So zero and π are not included. At zero and π, sine becomes zero. Since the boundaries are not included, we will not obtain f'(x) ≤ 0.
•
So f(x) is not just decreasing, it is strictly decreasing.
Part (b):
1. The interval that we need to consider in part (b) is (π,2π).
2. This interval indicates III and IV quadrants. In III and IV quadrants, sine is always −ve. So −sin x will be +ve.
3. We can write:
f'(x) > 0. So f(x) is strictly increasing in this interval.
4.
Note that (π,2π) is an open interval. So π and 2π are not included. At
π and 2π, sine becomes zero. Since the boundaries are not included,
we will not obtain f'(x) ≥ 0.
•
So f(x) is not just increasing, it is strictly increasing.
Part (c):
1. The interval that we need to consider in part (c) is (0,2π).
2. This interval indicates all the four quadrants. It includes both the intervals in parts (a) and (b).
3. So in (0,2π), f(x) is neither increasing nor decreasing.
•
Fig.22.5 below shows the graphs:
♦ f(x) = cos x (red curve)
♦ f'(x) = −sin x (yellow curve)
 |
| Fig.22.5 |
•
We see that:
♦ In the interval (0,π),
✰ yellow curve is −ve
✰ red curve is decreasing
♦ In the interval (π,2π),
✰ yellow curve is +ve
✰ red curve is increasing
Solved example 22.10
Find the intervals in which the function given by
f(x) = x2 − 4x + 6 is
(a) strictly increasing.
(b) strictly decreasing.
Solution:
First we write the derivative: f'(x) = 2x −4
Part (a):
1. For f(x) to be strictly increasing, f'(x) > 0.
2. So we can write:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{2x - 4} & {~>~} &{0} \\
{~\color{magenta} 2 } &{\implies} &{2x} & {~>~} &{4} \\
{~\color{magenta} 3 } &{\implies} &{x} & {~>~} &{2} \\
\end{array}$
(This is a linear inequality. We have seen how to solve such inequalities in chapter 6 of class 11. Details here)
3. That means, x can be any real number greater than 2. In such a situation, f(x) will be strictly increasing.
•
In other words, f(x) will be strictly increasing in the interval (2,∞).
•
We can represent this on a real number line as shown in fig.22.6(a) below:
 |
| Fig.22.6 |
4. Note that (2,∞) is an open interval. So 2 and ∞ are not included.
At 2, f'(x) becomes zero. Since the boundary 2 is not included,
we will not obtain f'(x) ≥ 0.
•
So f(x) is not just increasing, it is strictly increasing.
Part (b):
1. For f(x) to be strictly decreasing, f'(x) < 0.
2. So we can write:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{2x - 4} & {~<~} &{0} \\
{~\color{magenta} 2 } &{\implies} &{2x} & {~<~} &{4} \\
{~\color{magenta} 3 } &{\implies} &{x} & {~<~} &{2} \\
\end{array}$
3. That means, x can be any real number less than 2. In such a situation, f(x) will be strictly decreasing.
•
In other words, f(x) will be strictly decreasing in the interval (−∞,2).
•
We can represent this on a real number line as shown in fig.22.6(b) above.
4. Note that (−∞,2) is an open interval. So −∞ and 2 are not included.
At 2, f'(x) becomes zero. Since the boundary 2 is not included,
we will not obtain f'(x) ≤ 0.
•
So f(x) is not just decreasing, it is strictly decreasing.
•
Fig.22.7 below shows the graphs:
♦ f(x) = x2 − 4x + 6 (red curve)
♦ f'(x) = 2x −4 (yellow curve)
 |
| Fig.22.7 |
•
We see that:
♦ In the interval (−∞,2),
✰ yellow line is −ve
✰ red curve is decreasing
♦ In the interval (2,∞),
✰ yellow line is +ve
✰ red curve is increasing
♦ At x = 2, f'(x) = 0
✰ Here, f(x) is neither increasing, nor decreasing
Solved example 22.11
Find the intervals in which the function given by
f(x) = 4x3 − 6x2 − 72x + 30 is
(a) strictly increasing.
(b) strictly decreasing.
Solution:
First we write the derivative: f'(x) = 12x2 −12x − 72
= 12(x2 − x − 6)
= 12(x − 3)(x + 2)
Part (a):
1. For f(x) to be strictly increasing, f'(x) > 0.
2. So we can write:
12(x − 3)(x + 2) > 0
•
For this inequality to be true,
x must be either greater than 3 or less than −2
(In the previous problem, we obtained a linear inequality. It can be solved easily. But here, the inequality is not linear. We need to examine it carefully to find the appropriate x values)
•
x > 3 and x < −2 is possible only in two intervals:
(3,∞) and (−∞,−2)
•
We can represent this on a real number line as shown in fig.22.8(a) below:
 |
| Fig.22.8 |
3. Note that both are open intervals. So 3 and −2 are not included.
At those points, f'(x) becomes zero. Since they are not included,
we will not obtain f'(x) ≥ 0.
•
So f(x) is not just increasing, it is strictly increasing.
Part (b):
1. For f(x) to be strictly decreasing, f'(x) < 0.
2. So we can write:
12(x − 3)(x + 2) < 0
•
This is possible only in one interval:
(−2,3)
•
We can represent this on a real number line as shown in fig.22.8(b) above.
3. Note that it is an open interval. So −2 and 3 are not included.
At those points, f'(x) becomes zero. Since they are not included,
we will not obtain f'(x) ≤ 0.
•
So f(x) is not just decreasing, it is strictly decreasing.
•
Fig.22.9 below shows the graphs:
♦ f(x) = 4x3 − 6x2 − 72x + 30 (red curve)
♦ f'(x) = 12x2 −12x − 72 (yellow curve)
 |
| Fig.22.9 |
•
We see that:
♦ To the left of x = −2,
✰ yellow curve is +ve
✰ red curve is increasing
♦ Between x = −2 and x = 3,
✰ yellow curve is −ve
✰ red curve is decreasing
♦ To the right of x = 3,
✰ yellow curve is +ve
✰ red curve is increasing
♦ At x = −2, f'(x) = 0
✰ Here, f(x) is neither increasing, nor decreasing
♦ At x = 3, f'(x) = 0
✰ Here, f(x) is neither increasing, nor decreasing
Solved example 22.12
Find the intervals in which the function given by
$\rm{f(x)~=~\sin 3x, ~x \in \left[0,\frac{\pi}{2} \right]}$ is
(a) increasing
(b) decreasing
Solution:
First we write the derivative: f'(x) = 3 cos 3x
Part (a):
1. For f(x) to be increasing, f'(x) ≥ 0.
2. So we can write: 3 cos 3x ≥ 0
3. To solve this inequality, we will first write it as an equation. We get: 3 cos 3x = 0
This is possible only if cos 3x = 0.
4. This is a trigonometric equation. We will find the general solution. From the general solution, we can pickup the appropriate values.
For finding the general solution of cosine functions, we must use theorem 2 which we saw in chapter 3. Details here.
• For any real numbers x and y
cos x = cos y implies x = 2n𝞹 ± y, where n ∈ Z
• In the place of cos x, we have cos 3x.
• In the place of cos y, we have zero, which is cos (π/2).
So we can write:
cos 3x = cos (π/2) implies 3x = 2n𝞹 ± (π/2), where n ∈ Z
• This can be simplified as follows:
5. Now we can put various values of n. We get:
We see that:
• n = −1 and n = − 2 give x values out side the domain $\rm{ \left[0,\frac{\pi}{2} \right]}$
So we need not test for n values below −2
• n = 2 also gives x values out side the domain. So we need not test for n values above 2.
• Consider the following values
♦ $x = \rm{\frac{\pi}{6}}$ obtained when n = 0
♦ $x = \rm{\frac{\pi}{2}}$ obtained when n = 1
These are the only values which are present in the domain. The derivative becomes zero at these two values.
6. Let us mark the two acceptable values on the real number line. It is shown in fig.22.10 below:
 |
| Fig.22.10 |
•
A and B are the boundaries of the domain. Point C divides AB into two parts.
♦ In one of these two parts, f(x) will be increasing.
♦ In the other part, f(x) will be decreasing.
7. Let us consider an input x from between A and C. We get:
$\rm{0 \lt x \lt \frac{\pi}{6}}$
⇒ $\rm{0 \lt 3x \lt \frac{\pi}{2}}$
•
The input for cosine is "3x". This "3x" is increasing from 0 to $\rm{\frac{\pi}{2}}$.
•
So "3x" is in the quadrant I. In that quadrant, cosine is always +ve.
•
That means, if we take an input from between A and C, cos 3x will be greater than zero.
•
In other words, if we take an input from between A and C, f'(x) will be greater than zero.
•
Therefore, f(x) is strictly increasing in $\rm{\left(0,\frac{\pi}{6} \right)}$
8. Suppose that, the input is the exact point A.
Then we get: cos 3x = cos 0 = 1
Here also, f'(x) is greater than zero. So it is strictly increasing.
9. Suppose that, the input is the exact point C.
Then we get: $\rm{\cos 3x ~=~\cos \left(3 \times \frac{\pi}{6} \right)~=~\cos \frac{\pi}{2}~=~0}$
Here, f'(x) is equal to zero. So it is just increasing. Not strictly increasing.
10. Based on (7), (8) and (9), we can write:
♦ f(x) is strictly increasing on $\rm{\left[0,\frac{\pi}{6} \right)}$
♦ f(x) is increasing on $\rm{\left[0,\frac{\pi}{6} \right]}$
11. Let us consider an input x from between C and B. We get:
$\rm{\frac{\pi}{6} \lt x \lt \frac{\pi}{2}}$
⇒ $\rm{\frac{\pi}{2} \lt 3x \lt \frac{3 \pi}{2}}$
•
The input for cosine is "3x". This "3x" is increasing from $\rm{\frac{\pi}{2}}$ to $\rm{\frac{3 \pi}{2}}$.
•
So "3x" is in the quadrants II and III. In those quadrants, cosine is always −ve.
•
That means, if we take an input from between C and B, cos 3x will be less than zero.
•
In other words, if we take an input from between C and B, f'(x) will be less than zero.
•
Therefore, f(x) is strictly decreasing in $\rm{\left( \frac{\pi}{6} , \frac{\pi}{2} \right)}$
12. Suppose that, the input is the exact point C.
•
Then we get: $\rm{\cos 3x \,=\, \cos \left(3 \times \frac{\pi}{6} \right) \,=\, \cos \frac{\pi}{2} \,=\,0}$
•
Here, f'(x) is equal to zero. So it is just decreasing, Not strictly decreasing.
13. Suppose that, the input is the exact point B.
•
Then we get: $\rm{\cos 3x ~=~\cos \left(3 \times \frac{\pi}{2} \right)~=~\cos \frac{3 \pi}{2}~=~0}$
•
Here, f'(x) is equal to zero. So it is just decreasing, Not strictly decreasing.
14. Based on (11), (12) and (13), we can write:
♦ f(x) is strictly decreasing in $\rm{\left( \frac{\pi}{6} , \frac{\pi}{2} \right)}$
♦ f(x) is increasing on $\rm{\left[ \frac{\pi}{6} , \frac{\pi}{2} \right]}$
•
Fig.22.11 below shows the graphs:
♦ f(x) = sin 3x
♦ f'(x) = 3 cos 3x
 |
| Fig.22.11 |
•
We see that:
♦ Between A and C,
✰ yellow curve is +ve
✰ red curve is increasing
♦ Between C and B,
✰ yellow curve is −ve
✰ red curve is decreasing
♦ At x = π/6, f'(x) = 0
✰ Here, f(x) is neither increasing, nor decreasing
♦ At x = π/2, f'(x) = 0
✰ Here, f(x) is neither increasing, nor decreasing
Solved example 22.13
Find the intervals in which the function given by
$\rm{f(x)~=~\sin x \,+\, \cos x, ~x \in \left[0,2 \pi \right]}$ is
(a) Strictly increasing
(b) Strictly decreasing
Solution:
First we write the derivative: f'(x) = cos x − sin x
1. For f(x) to be increasing, f'(x) > 0.
2. So we can write: cos x − sin x > 0
3. To solve this inequality, we will first write it as an equation. We get: cos x − sin x = 0
⇒ cos x = sin x
⇒ cos x = cos (π/2 − x)
4. This is a trigonometric equation. We will find the general solution. From the general solution, we can pickup the appropriate values.
For finding the general solution of cosine functions, we must use theorem 2 which we saw in chapter 3. Details here.
• For any real numbers x and y
cos x = cos y implies x = 2n𝞹 ± y, where n ∈ Z
• In the place of cos x, we have cos x.
• In the place of cos y, we have cos (π/2 − x).
So we can write:
cos x = cos (π/2 − x) implies x = 2n𝞹 ± (π/2 − x), where n ∈ Z
• This can be simplified as follows:
5. Now we can put various values of n. We get:
We see that:
• n = −1 and n = − 2 give x values out side the domain $\rm{ \left[0,2 \pi \right]}$
So we need not test for n values below −2
• n = 2 also gives x values out side the domain. So we need not test for n values above 2.
• Consider the following values
♦ $x = \rm{\frac{\pi}{4}}$ obtained when n = 0
♦ $x = \rm{\frac{5 \pi}{4}}$ obtained when n = 1
These are the only values which are present in the domain. The derivative becomes zero at these two values.
6. Let us mark the two acceptable values on the real number line. It is shown in fig.22.12 below:
 |
| Fig.22.12 |
• A and B are the boundaries of the domain. Points C and D divide AB into three parts. AC, CD and DB. We need to analyze each of the three parts.
7. For easy analysis, it is better to rearrange f'(x) as the product of two trigonometric functions. This can be done as follows:
$\begin{array}{ll} &{\color{magenta}1} &{} &{f’(x)} &{}={} &{\cos x - \sin x} &{} \\
&{~\color{magenta}2} &{} &{{}} &{}={} &{\cos x - \cos \left(\frac{\pi}{2} – x \right)} &{} \\
&{~\color{magenta}3} &{} &{{}} &{}={} &{-2 \sin \left(\frac{x + \pi/2 - x}{2} \right) \sin \left(\frac{x - \pi/2 + x}{2} \right)} &{} \\
&{~\color{magenta}4} &{} &{{}} &{}={} &{-2 \sin \left(\frac{\pi}{4} \right) \sin \left(x - \frac{\pi}{4} \right)} &{} \\
&{~\color{magenta}5} &{} &{{}} &{}={} &{-2 \frac{1}{\sqrt 2} \sin \left(x - \frac{\pi}{4} \right)} &{} \\
&{~\color{magenta}6} &{} &{{}} &{}={} &{-{\sqrt 2} \sin \left(x - \frac{\pi}{4} \right)} &{} \\
\end{array}$
8. Let us consider the first interval AC.
• When the input x from between A and C, x will be less than $\frac{\pi}{4}$
• So $(x - \frac{\pi}{4})$ will be −ve.
$\implies \sin(x - \frac{\pi}{4})$ is −ve.
$\implies -\sqrt{2}\sin(x - \frac{\pi}{4})$ is +ve`.
9. Suppose that, the input is the exact point A.
Then we get:
$\begin{array}{ll} &{\color{magenta}1} &{} &{f’(x)} &{}={} &{-\sqrt{2}\sin(x - \frac{\pi}{4})} &{} \\
&{~\color{magenta}2} &{} &{{}} &{}={} &{-\sqrt{2}\sin(0 - \frac{\pi}{4})} &{} \\
&{~\color{magenta}3} &{} &{{}} &{}={} &{-\sqrt{2}\sin(- \frac{\pi}{4})} &{} \\
&{~\color{magenta}4} &{} &{{}} &{}={} &{-\sqrt{2} . -\sqrt{2}} &{} \\
&{~\color{magenta}5} &{} &{{}} &{}={} &{\sqrt{2}} &{} \\
\end{array}$
• Here also, f'(x) is +ve. So it is strictly increasing.
10. Suppose that, the input is the exact point C.
Then we get:
$\begin{array}{ll} &{\color{magenta}1} &{} &{f’(x)} &{}={} &{-\sqrt{2}\sin(x - \frac{\pi}{4})} &{} \\
&{~\color{magenta}2} &{} &{{}} &{}={} &{-\sqrt{2}\sin(\frac{\pi}{4} - \frac{\pi}{4})} &{} \\
&{~\color{magenta}3} &{} &{{}} &{}={} &{-\sqrt{2}\sin(0)} &{} \\
&{~\color{magenta}4} &{} &{{}} &{}={} &{0} &{} \\
\end{array}$
• f'(x) is zero. So at C, the given function is just increasing. Not strictly increasing.
11. We are asked to find the intervals where the function is strictly increasing or strictly decreasing. Based on (8), (9) and (10), we can write:
♦ f(x) is strictly increasing on $\rm{\left[0,\frac{\pi}{4} \right)}$
12. Let us consider the second interval CD.
• When the input x from between C and D, x will be between $\frac{\pi}{4}$ and $\frac{5 \pi}{4}$
• So $(x - \frac{\pi}{4})$ will be between zero and π. It is the I and II quadrants.
$\implies \sin(x - \frac{\pi}{4})$ is +ve.
$\implies -\sqrt{2}\sin(x - \frac{\pi}{4})$ is −ve.
13. Suppose that, the input is the exact point C.
• This is same as (10) above.
• f'(x) is zero. So at C, the given function is just decreasing. Not strictly decreasing.
14. Suppose that, the input is the exact point D.
Then we get:
\begin{array}{ll} &{\color{magenta}1} &{} &{f’(x)} &{}={} &{-\sqrt{2}\sin(x - \frac{\pi}{4})} &{} \\
&{~\color{magenta}2} &{} &{{}} &{}={} &{-\sqrt{2}\sin(\frac{5 \pi}{4} - \frac{\pi}{4})} &{} \\
&{~\color{magenta}3} &{} &{{}} &{}={} &{-\sqrt{2}\sin(\pi)} &{} \\
&{~\color{magenta}4} &{} &{{}} &{}={} &{0} &{} \\
\end{array}
• Here also, f'(x) is zero. So at D, the given function is just decreasing. Not strictly decreasing.
15. Based on (12), (13) and (14), we can write:
♦ f(x) is strictly decreasing in $\rm{\left( \frac{\pi}{4} , \frac{5 \pi}{4} \right)}$
16. Let us consider the last interval DB.
• When the input x from between D and B, x will be between $\frac{5 \pi}{4}$ and 2π.
• So $(x - \frac{\pi}{4})$ will be between π and $\frac{7 \pi}{4}$. It is the III and IV quadrants.
$\implies \sin(x - \frac{\pi}{4})$ is −ve.
$\implies -\sqrt{2}\sin(x - \frac{\pi}{4})$ is +ve.
17. Suppose that, the input is the exact point D.
• This is same as (14) above.
• f'(x) is zero. So at D, the given function is just increasing. Not strictly increasing.
18. Suppose that, the input is the exact point B.
Then we get:
\begin{array}{ll} &{\color{magenta}1} &{} &{f’(x)} &{}={} &{-\sqrt{2}\sin(x - \frac{\pi}{4})} &{} \\
&{~\color{magenta}2} &{} &{{}} &{}={} &{-\sqrt{2}\sin(2 \pi - \frac{\pi}{4})} &{} \\
&{~\color{magenta}3} &{} &{{}} &{}={} &{-\sqrt{2} \left( -\sin \frac{\pi}{4} \right)} &{} \\
&{~\color{magenta}4} &{} &{{}} &{}={} &{-\sqrt{2} . -\sqrt{2}} &{} \\
&{~\color{magenta}5} &{} &{{}} &{}={} &{2} &{} \\
\end{array}
• f'(x) is +ve. So at D, the given function is strictly increasing.
19. Based on (16), (17) and (18), we can write:♦ f(x) is strictly increasing on $\rm{\left(\frac{5 \pi}{4}, 2 \pi \right]}$
•
Fig.22.11 below shows the graphs:
♦ f(x) = sin x + cos x (red color)
♦ f'(x) = cos x − sin x (yellow color)
 |
| Fig.22.13 |
•
We see that:
♦ Between A and C,
✰ yellow curve is +ve
✰ red curve is increasing
♦ Between C and D,
✰ yellow curve is −ve
✰ red curve is decreasing
♦ Between D and B,
✰ yellow curve is +ve
✰ red curve is increasing
♦ At x = C, f'(x) = 0
✰ Here, f(x) is neither increasing, nor decreasing
♦ At x = D, f'(x) = 0
✰ Here, f(x) is neither increasing, nor decreasing
The link below gives a few more solved examples:
In the next section, we will see tangents and normals.
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