Question

Using differentials, find the intervals of increasing and decreasing for

$f(x) = {(x - 1)}^{2} (x - 3)^{2}$
​

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = {(x - 1)}^{2} {(x - 3)}^{2}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx} {(x - 1)}^{2} {(x - 3)}^{2}$

We know,

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

So, using this, we get

$\rm :\longmapsto\:f'(x) = {(x - 1)}^{2}\dfrac{d}{dx} {(x - 3)}^{2} + {( x - 3)}^{2} \dfrac{d}{dx} {(x - 1)}^{2}$

We know,

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

So, using this, we get

$\rm :\longmapsto\:f'(x) = {(x - 1)}^{2} \times 2(x - 3) + {(x - 3)}^{2} \times 2(x - 1)$

$\rm :\longmapsto\:f'(x) = 2(x - 1)(x - 3)(x - 1 + x - 3)$

$\rm :\longmapsto\:f'(x) = 2(x - 1)(x - 3)(2x - 4)$

$\rm :\longmapsto\:f'(x) = 4(x - 1)(x - 3)(x - 2)$

So, critical points are

$\rm :\longmapsto\:\boxed{ \tt{ \: x = 1, \: 2, \: 3 \: }}$

Now, Let we check the sign of f'(x) in the intervals.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ( - \infty ,1) & \sf - ve \\ \\ \sf (1,2) & \sf + ve \\ \\ \sf (2,3) & \sf - ve\\ \\ \sf (3, \infty ) & \sf + ve \end{array}} \\ \end{gathered}$

So,

For increasing,

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

$\bf\implies \:x \: \in \: (1,2) \: \cup \: (3, \infty )$

And

For decreasing

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

$\bf\implies \:x \: \in \: ( - \infty ,1) \: \cup \: (2,3)$

More to know :-

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