Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which $f\left(x\right) = \dfrac{x}{\log x}$ is increasing or decreasing.

$f\left(x\right) = \dfrac{x}{\log x}$

$\therefore$ $f'\left(x\right) = \dfrac{\log x - x \times \dfrac{1}{x}}{\left(\log x\right)^2}$

i.e. $f'\left(x\right) = \dfrac{\log x -1}{\left(\log x\right)^2}$

Now, $\left(\log x\right)^2 > 0 \;\; \forall \; x $

$\therefore$ $f'\left(x\right) > 0$ $\implies$ $\log x - 1 > 0$

i.e. $\log x > 1$ $\implies$ $\log x > \log e$ $\implies$ $x>e$

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left(e,\infty\right)$

$f'\left(x\right) < 0$ $\implies$ $\log x -1 < 0$

i.e. $\log x < 1$ $\implies$ $x < e$

$\therefore$ $f\left(x\right)$ is decreasing in the interval $\left(0,e\right)$

Application of Derivatives: Increasing Decreasing Functions

Determine the intervals in which $f:R \rightarrow R$, $f\left(x\right) = 2 \cos x + \sin^2 x$ is strictly increasing or strictly decreasing.

$f\left(x\right) = 2 \cos x + \sin^2 x$

$\therefore$ $f'\left(x\right) = -2 \sin x + 2 \sin x \cos x$

i.e. $f'\left(x\right) = -2 \sin x \left(1 - \cos x\right) $

i.e. $f'\left(x\right) = -2 \sin x \times 2 \sin^2 \left(\dfrac{x}{2}\right)$

i.e. $f'\left(x\right) = -4 \sin x \sin^2 \left(\dfrac{x}{2}\right)$ $\;\; \cdots$ (1)

Now, $\sin^2 \left(\dfrac{x}{2}\right) > 0 \;\; \forall \; x \in R$

When $f'\left(x\right) > 0$ $\implies$ $f\left(x\right)$ is increasing.

Now, from equation (1), $f'\left(x\right) > 0$ $\implies$ $\sin x < 0$

$\implies$ $- \pi < x < 0$

Generalizing, we have $\left(2 k - 1\right)\pi < x < 2 k \pi$

i.e. $f\left(x\right)$ is strictly increasing in the interval $\left(\left(2 k - 1\right)\pi, 2 k \pi\right), \;\; k \in Z$

When $f'\left(x\right) < 0$ $\implies$ $f\left(x\right)$ is decreasing.

Now, from equation (1), $f'\left(x\right) < 0$ $\implies$ $\sin x > 0$

$\implies$ $0 < x < \pi$

Generalizing, we have $2 k \pi < x < \left(2 k + 1\right) \pi$

i.e. $f\left(x\right)$ is strictly decreasing in the interval $\left(2 k \pi, \left(2 k + 1 \right) \pi \right), \;\; k \in Z$

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f\left(x\right) = \dfrac{4x^2 + 1}{x}$ is increasing and the intervals in which it is decreasing.

$f\left(x\right) = \dfrac{4x^2 + 1}{x}$ $\implies$ $f\left(x\right) = 4x + \dfrac{1}{x}$

$\therefore$ $f'\left(x\right) = 4 - \dfrac{1}{x^2} = \dfrac{4x^2 -1}{x^2}$

Now, $x^2 > 0 \;\; \forall \; x, \;\; x \neq 0$

$f'\left(x\right) = 0$ $\implies$ $4x^2 -1 = 0$

i.e. $x^2 = \dfrac{1}{4}$ $\implies$ $x = \dfrac{1}{2}$, $x = -\dfrac{1}{2}$

These points divide the real line into three intervals namely $\left(-\infty, -\dfrac{1}{2}\right)$, $\left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$ and $\left(\dfrac{1}{2},\infty\right)$

Interval	Sign of $\left(4x^2 - 1\right)$	Sign of $f'\left(x\right)$	Nature of function f
$\left(-\infty,-\dfrac{1}{2}\right)$	+ve	+ve	Increasing
$\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$	-ve	-ve	Decreasing
$\left(\dfrac{1}{2},\infty\right)$	+ve	+ve	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left\{x : x < \dfrac{-1}{2}\right\} \cup \left\{x : x > \dfrac{1}{2}\right\} $

$f\left(x\right)$ is decreasing in the interval $\left\{x : \dfrac{-1}{2} < x < \dfrac{1}{2}\right\} $

Application of Derivatives: Increasing Decreasing Functions

Determine the intervals in which $f\left(x\right) = \sin^4 x + \cos^4 x$, $x \in \left(0,\dfrac{\pi}{2}\right)$ is increasing and in which it is decreasing.

$f\left(x\right) = \sin^4 x + \cos^4 x$

$\therefore$ $f'\left(x\right) = 4 \sin^3 x \cos x - 4 \cos^3 x \sin x$

i.e. $f'\left(x\right) = 4 \sin x \cos x \left(\sin^2 x - \cos^2 x\right)$

i.e. $f'\left(x\right) = -4 \sin x \cos x \cos 2x$ $\;\; \cdots$ (1)

Now, $\sin x > 0 \; \forall \; x \in \left(0,\dfrac{\pi}{2}\right)$; $\cos x > 0 \; \forall \; x \in \left(0,\dfrac{\pi}{2}\right)$

$f\left(x\right)$ is increasing when $f'\left(x\right) > 0$

Since $\sin x > 0$ and $\cos x > 0$, $\therefore$ $\;$ $f'\left(x\right) > 0$ $\implies$ $\cos 2x < 0$

i.e. $\dfrac{\pi}{2} < 2x < \pi$ or $\pi < 2x < \dfrac{3\pi}{2}$

i.e. $\dfrac{\pi}{4} < x < \dfrac{\pi}{2}$ or $\dfrac{\pi}{2} < x < \dfrac{3\pi}{4}$

Since $x \in \left(0,\dfrac{\pi}{2}\right)$, therefore $f\left(x\right)$ is increasing in $\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$

$f\left(x\right)$ is decreasing when $f'\left(x\right) < 0$

Since $\sin x > 0$ and $\cos x > 0$, $\therefore$ $\;$ $f'\left(x\right) < 0$ $\implies$ $\cos 2x > 0$

i.e. $0 < 2x < \dfrac{\pi}{2}$ or $\dfrac{3\pi}{2} < 2x < 2\pi$

i.e. $0 < x < \dfrac{\pi}{4}$ or $\dfrac{3\pi}{4} < x < \pi $

Since $x \in \left(0,\dfrac{\pi}{2}\right)$, therefore $f\left(x\right)$ is decreasing in $\left(0,\dfrac{\pi}{4}\right)$

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f\left(x\right) = \sin x - \cos x$, $0 < x < 2 \pi$ is increasing or decreasing.

$f\left(x\right) = \sin x - \cos x$

$\begin{aligned} \therefore \; f'\left(x\right) & = \cos x + \sin x \\ & = \sqrt{2}\left(\dfrac{1}{\sqrt{2}} \cos x + \dfrac{1}{\sqrt{2}} \sin x\right) \\ & = \sqrt{2} \left(\sin \dfrac{\pi}{4} \cos x + \cos \dfrac{\pi}{4} \sin x\right) \\ & = \sqrt{2} \sin \left(\dfrac{\pi}{4} + x\right) \end{aligned}$

$\therefore$ $\;$ $f'\left(x\right) = 0$ $\implies$ $\sin \left(\dfrac{\pi}{4} + x\right) = 0$

i.e. $\sin \left(\dfrac{\pi}{4} + x\right) = \sin 0 \;\; \text{OR} \;\; \sin \pi \;\; \text{OR} \;\; \sin 2 \pi$

$\implies$ $x = - \dfrac{\pi}{4}$ OR $x = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$ OR $x = 2 \pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4}$

Since $0 < x < 2 \pi$, $\;$ $x = - \dfrac{\pi}{4}$ is not a valid solution.

The points $x = \dfrac{3\pi}{4}$ and $x = \dfrac{7\pi}{4}$ divide the interval $0 < x < 2 \pi$ into three intervals namely $\left(0,\dfrac{3\pi}{4}\right)$, $\left(\dfrac{3\pi}{4},\dfrac{7\pi}{4}\right)$ and $\left(\dfrac{7\pi}{4},2\pi\right)$

Interval	Sign of $\sin \left(\dfrac{\pi}{4} + x\right)$	Nature of function f
$\left(0,\dfrac{3\pi}{4}\right)$	+ve	Increasing
$\left(\dfrac{3\pi}{4},\dfrac{7\pi}{4}\right)$	-ve	Decreasing
$\left(\dfrac{7\pi}{4}, 2 \pi\right)$	+ve	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left\{x : 0< x < \dfrac{3\pi}{4}\right\} \cup \left\{x : \dfrac{7\pi}{4}< x < 2\pi\right\} $

$f\left(x\right)$ is decreasing in the interval $\left\{x : \dfrac{7\pi}{4} < x < 2\pi\right\} $

Application of Derivatives: Increasing Decreasing Functions

Find the values of x for which $y = \left[x\left(x-2\right)\right]^2$ is an increasing function.

$y = \left[x\left(x-2\right)\right]^2$

$\therefore$ $\dfrac{dy}{dx}= 2 x \left(x-2\right) \left(x+x-2\right)$

i.e. $\dfrac{dy}{dx}= 4x \left(x-2\right) \left(x-1\right)$

$\dfrac{dy}{dx}=0$ $\implies$ $x=0$, $x=2$, $x=1$

These points divide the real line into four disjoint intervals namely $\left(-\infty,0\right)$, $\left(0,1\right)$, $\left(1,2\right)$ and $\left(2,\infty\right)$

Interval	Sign of $\dfrac{dy}{dx}$	Nature of function y
$\left(-\infty,0\right)$	-ve	Decreasing
$\left(0,1\right)$	+ve	Increasing
$\left(1,2\right)$	-ve	Decreasing
$\left(2,\infty\right)$	+ve	Increasing

$\therefore$ y is increasing in the intervals $\left(0,1\right)$ and $\left(2,\infty\right)$.

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f\left(x\right) = \left(x+1\right)^2 \left(x-3\right)^3$ is increasing or decreasing.

$f\left(x\right) = \left(x+1\right)^2 \left(x-3\right)^3$

$\therefore$ $f'\left(x\right) = 3 \left(x+1\right)^2 \left(x-3\right)^2 + 2 \left(x+1\right) \left(x-3\right)^3$

i.e. $f'\left(x\right) = \left(x+1\right) \left(x-3\right)^2 \left[3\left(x+1\right) +2 \left(x-3\right)\right]$

i.e. $f'\left(x\right) = \left(x+1\right) \left(x-3\right)^2 \left(5x-3\right)$

Now, $\left(x-3\right)^2 > 0 \; \forall \; x$

For critical points, $f'\left(x\right) = 0$

$\implies$ $x+1 = 0$ $\implies$ $x=-1$

or $\left(x-3\right)^2 =0$ $\implies$ $x=3$

or $5x-3=0$ $\implies$ $x=\dfrac{3}{5}$

These points divide the real line into four intervals namely $\left(-\infty, -1\right)$, $\left(-1, \dfrac{3}{5}\right)$, $\left(\dfrac{3}{5},3\right)$ and $\left(3,\infty\right)$

Interval	Sign of $\left(x+1\right)$	Sign of $\left(5x-3\right)$	Sign of $f'\left(x\right)$	Nature of function f
$\left(-\infty,-1\right)$	-ve	-ve	+ve	Increasing
$\left(-1,\dfrac{3}{5}\right)$	+ve	-ve	-ve	Decreasing
$\left(\dfrac{3}{5},3\right)$	+ve	+ve	+ve	Increasing
$\left(3,\infty\right)$	+ve	+ve	+ve	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left\{x : x<-1\right\} \cup \left\{x : x>\dfrac{3}{5}\right\} $

$f\left(x\right)$ is decreasing in the interval $\left\{x : -1 < x < \dfrac{3}{5}\right\} $

Application of Derivatives: Increasing Decreasing Functions

Find the least value of $a$ such that the function f given by $f\left(x\right) = x^2 +ax + 1$ is strictly increasing on $\left(1,2\right)$.

$f\left(x\right) = x^2 + ax + 1$

i.e. $f'\left(x\right) = 2x + a$

Since $f\left(x\right)$ is a strictly increasing function $\implies$ $f'\left(x\right) > 0$

i.e. $2 x + a > 0$ $\implies$ $x > -\dfrac{a}{2}$

i.e. To find $x > \dfrac{-a}{2}$ when x is in the interval $\left(1,2\right)$

i.e. To find $x > \dfrac{-a}{2}$ when $1 < x < 2$

Least value of x is 1.

$\therefore$ Least value of a is obtained when $x = 1$

$\therefore$ We have, $1 = \dfrac{-a}{2}$ $\implies$ $a = -2$

$\therefore$ The least value of $a = -2$.

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f:R \rightarrow R$, $f\left(x\right) = \dfrac{4 \sin x -2x - x \cos x}{2+\cos x}$ is increasing and in which it is decreasing.

$f\left(x\right) = \dfrac{4 \sin x -2x - x \cos x}{2 + \cos x}$

i.e. $f\left(x\right) = \dfrac{4\sin x}{2+\cos x} - \dfrac{x \left(2+\cos x\right)}{2+\cos x}$

i.e. $f\left(x\right) = \dfrac{4\sin x}{2+\cos x} - x$

$\therefore$ $f'\left(x\right) = \dfrac{\left(2+\cos x\right)\times 4 \cos x -4 \sin x \times \left(-\sin x\right)}{\left(2+\cos x\right)^2} -1$

i.e. $f'\left(x\right) = \dfrac{8 \cos x + 4 \cos ^2 x + 4 \sin ^2 x}{\left(2+\cos x\right)^2}-1$

i.e. $f'\left(x\right) = \dfrac{8 \cos x + 4 -4 - 4\cos x -\cos^2 x}{\left(2+\cos x\right)^2}$

i.e. $f'\left(x\right) = \dfrac{4 \cos x - \cos^2 x}{\left(2+\cos x\right)^2}$

i.e. $f'\left(x\right) = \dfrac{\cos x \left(4- \cos x\right)}{\left(2+\cos x\right)^2}$ $\;\; \cdots$ (1)

In equation (1), $\left(2+\cos x\right)^2 > 0 \; \forall \; x$ $\;\; \cdots$ (2)

Also, $4 - \cos x > 0 \; \forall \; x$ since $\cos x$ varies between $\pm 1$ $\;\; \cdots$ (3)

Case (i): $\mathbf{f\left(x\right)}$ is increasing:

$f\left(x\right)$ is increasing when $f'\left(x\right) > 0$

$\therefore$ From equation (1), for $f'\left(x\right) > 0$, $\cos x > 0$

Now, $\cos x > 0$ i.e. positive in the First and Fourth quadrants.

$\therefore$ $\cos x > 0$ $\implies$ $0 < x < \dfrac{\pi}{2}$ (First Quadrant) and $\dfrac{3\pi}{2} < x < 2\pi$ (Fourth Quadrant)

$\therefore$ Generalizing,

$\cos x > 0$ $\implies$ $2k\pi < x < \left(4k+1\right)\dfrac{\pi}{2}$ (First Quadrant) and $\left(4k+3\right) \dfrac{\pi}{2} < x < \left(2k+2\right)\pi$ (Fourth Quadrant), $k \in Z$

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left(2k\pi, \left(4k+1\right)\dfrac{\pi}{2}\right)$ and $\left(\left(4k+3\right)\dfrac{\pi}{2}, \left(2k+2\right)\pi\right)$, $k \in Z$

Case (ii): $\mathbf{f\left(x\right)}$ is decreasing:

$f\left(x\right)$ is decreasing when $f'\left(x\right) < 0$

$\therefore$ In view of equations (2) and (3), we have from equation (1), for $f'\left(x\right) < 0$, $\cos x < 0$

Now, $\cos x < 0$ i.e. negative in the Second and Third quadrants.

$\therefore$ $\cos x < 0$ $\implies$ $\dfrac{\pi}{2} < x < \pi$ (Second Quadrant) and $\pi < x < \dfrac{3\pi}{2}$ (Third Quadrant)

$\therefore$ Generalizing,

$\cos x < 0$ $\implies$ $\left(4k+1\right)\dfrac{\pi}{2} < x < \left(2k+1\right)\pi$ (Second Quadrant) and $\left(2k+1\right)\pi < x < \left(4k+3\right) \dfrac{\pi}{2}$ (Third Quadrant), $k \in Z$

$\therefore$ $f\left(x\right)$ is decreasing in the interval $\left(\left(4k+1\right)\dfrac{\pi}{2}, \left(2k+1\right)\pi \right)$ and $\left(\left(2k+1\right) \pi, \left(4k+3\right)\dfrac{\pi}{2}\right)$, $k \in Z$

Application of Derivatives: Increasing Decreasing Functions

Discuss the nature of the function f given by $f\left(x\right) = \log \left[\sin \left(x\right)\right]$ in the intervals $\left(0,\dfrac{\pi}{2}\right)$ and $\left(\dfrac{\pi}{2}, \pi\right)$.

$f\left(x\right) = \log \left[\sin \left(x\right)\right]$

$\therefore$ $f'\left(x\right) = \dfrac{\cos x}{\sin x} = \cot x$

For $0 < x < \dfrac{\pi}{2}$, $\cot x > 0$

i.e. $f'\left(x\right) > 0$ for $0 < x < \dfrac{\pi}{2}$

i.e. $f\left(x\right)$ is strictly increasing in the interval $\left(0,\dfrac{\pi}{2}\right)$

For $\dfrac{\pi}{2} < x < \pi$, $\cot x < 0$

i.e. $f'\left(x\right) < 0$ for $\dfrac{\pi}{2} < x < \pi$

i.e. $f\left(x\right)$ is strictly decreasing in the interval $\left(\dfrac{\pi}{2}, \pi\right)$

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which $f\left(x\right) = \dfrac{3}{10}x^4-\dfrac{4}{5}x^3 -3x^2 + \dfrac{36}{5}x + 11$ is increasing and in which it is decreasing.

$f\left(x\right) = \dfrac{3}{10}x^4 - \dfrac{4}{5}x^3 - 3x^2 + \dfrac{36}{5}x + 11$

$\therefore$ $f'\left(x\right) = \dfrac{6}{5}x^3-\dfrac{12}{5}x^2-6x+\dfrac{36}{5}$

i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x^3 -2x^2-5x+6\right)$

i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x-1\right) \left(x^2 -x-6\right)$

i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x-1\right) \left(x-3\right) \left(x+2\right)$

$\therefore$ For critical points, $f'\left(x\right) = 0$

$\implies$ $\left(x-1\right) \left(x-3\right) \left(x+2\right) = 0$

$\implies$ $x = -2$, $x=1$, $x=3$

These points divide the real line into four disjoint intervals namely $\left(-\infty,-2\right)$, $\left(-2,1\right)$, $\left(1,3\right)$ and $\left(3,\infty\right)$

Interval	Sign of $f'\left(x\right)$	Nature of function f
$\left(-\infty,-2\right)$	$< 0$	Decreasing
$\left(-2,1\right)$	$> 0$	Increasing
$\left(1,3\right)$	$ < 0$	Decreasing
$\left(3,\infty\right)$	$ > 0$	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the intervals $\left(-2,1\right)$ and $\left(3,\infty\right)$

$f\left(x\right)$ is decreasing in the intervals $\left(-\infty,-2\right)$ and $\left(1,3\right)$

Application of Derivatives: Increasing Decreasing Functions

Prove that the function given by $f\left(x\right) = x^3 - 3x^2 +3x - 100$ is increasing in $\mathbb{R}$.

$f\left(x\right) = x^3 -3x^2 + 3x -100$

$\begin{aligned} \therefore f'\left(x\right) & = 3x^2 -6x +3 \\ & = 3 \left(x^2 -2x +1\right) \\ & = 3 \left(x-1\right)^2 \end{aligned}$

Now, $\left(x-1\right)^2 > 0 \; \forall \; \mathbb{R}$

$\therefore$ $3 \left(x-1\right)^2 > 0 \; \forall \; \mathbb{R}$

$\implies$ $f'\left(x\right) > 0 \; \forall \; \mathbb{R}$

$\implies$ $f\left(x\right)$ is strictly increasing in $\mathbb{R}$.