Question

Find the absolute maxima and absolute minima of the function f(x)=x^2-1 on the interval [-1, 2]​

Answer 1

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

Given function is

$\rm \: f(x) \: = \: {x}^{2} - 1 \: \: \: \: x \in \: [ - 1,2]$

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

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

We know,

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

So, using this, we get

$\rm \: f'(x) = 2x$

For maxima and minima

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

$\rm \: 2x = 0$

$\rm\implies \:x = 0$

So, critical points are

$\rm \: x = - 1,0,2 \: \in \: \: [ - 1,2]$

Now, lets evaluate the value of f(x) at critical points.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf f(x) = {x}^{2} - 1 \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - 1 & \sf 0 \\ \\ \sf 0 & \sf - 1 \\ \\ \sf 2 & \sf 3 \end{array}} \\ \end{gathered}$

$\rm\implies \:f(x) \: is \: minimum \: at \: x = 0 \: and \: min \: value \: is \: - 1$

and

$\rm\implies \:f(x) \: is \: maximum \: at \: x = 2 \: and \: max \: value \: is \: 3$

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Additional Information  :-

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

To find the local 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.