Question

Find the intervals in which the following functions are increasing or
decreasing : f(x)= (x-1)(x-2)^2​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = (x - 1) {(x - 2)}^{2}$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) =\dfrac{d}{dx} (x - 1) {(x - 2)}^{2}$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}uv = u\dfrac{d}{dx}v + v\dfrac{d}{dx}u \: }}$

So, using this, we get

$\rm :\longmapsto\:f'(x) = (x - 1)\dfrac{d}{dx} {(x - 2)}^{2} + {(x - 2)}^{2}\dfrac{d}{dx}(x - 1)$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

So, using this, we get

$\rm :\longmapsto\:f'(x) = (x - 1)2(x - 2) + {(x - 2)}^{2}(1 - 0)$

$\rm :\longmapsto\:f'(x) =2 (x - 1)(x - 2) + {(x - 2)}^{2}$

$\rm :\longmapsto\:f'(x) =(x - 2)[2x - 2 + x - 2]$

$\rm :\longmapsto\:f'(x) =(x - 2)[3x - 4]$

So, critical points are

$\rm :\longmapsto\:\boxed{ \tt{ \: x = \frac{4}{3} \: \: and \: \: x = 2 \: }}$

Let we check the sign of f'(x).

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Interval & \bf Sign \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \bigg( - \infty ,\dfrac{4}{3} \bigg) & \sf + ve \\ \\ \sf \bigg(\dfrac{4}{3} ,2 \bigg) & \sf - ve \\ \\ \sf (2, \infty ) & \sf + ve \end{array}} \\ \end{gathered}$

So, for increasing

$\rm :\longmapsto\:f'(x) > 0$

$\bf\implies \:x \: \in \: \bigg( - \infty ,\dfrac{4}{3} \bigg) \: \cup \: (2, \: \infty )$

And

For decreasing,

$\rm :\longmapsto\:f'(x) < 0$

$\bf\implies \: \: x \: \in \: \bigg(\dfrac{4}{3} , \: 2 \bigg)$

More to know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$