Question

Find local maxima and minima for f(x) = x³.
Also find it's point of inflection. ​

Answer 1

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

Give function is

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

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

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

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

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

For maxima or minima

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

$\rm :\longmapsto\: {3x}^{2} = 0$

$\rm\implies \:x = 0$

Now, we have

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

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

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

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

$\rm :\longmapsto\:f''(x) =6x$

So,

$\rm :\longmapsto\:f''(0) =6 \times 0 = 0$

$\rm\implies \:0 \: is \: the \: point \: of \: inflection$

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Basic Concept Used :-

Let y = f(x) be a given function.

To find the maximum and minimum value, the following steps are follow :

1. Differentiate the given function.

2. For maxima or minima, put f'(x) = 0 and find critical points.

3. Then find the second derivative, i.e. f''(x).

4. Apply the critical points ( evaluated in second step ) in the second derivative.

5. Condition :-

The function f (x) is maximum when f''(x) < 0.

The function f (x) is minimum when f''(x) > 0.

The function f (x) has point of inflection when f''(x) = 0.

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EXPLORE MORE

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