Question

How do you find the relative extrema for $f(x) = 2x - 3{x^{\dfrac{2}{3}}} + 2$ on the interval $[ - 1,3]$ ?

Answer 1

Hint : First of all, extrema here refers to the maxima which we study in differential calculus. The maxima of any function can be calculated by finding its critical points the critical points are of two types:
A.Where $f'(x)$ becomes equal to zero
B.Where $f'(x)$ becomes not defined.
Here both types of critical points are present first we will differentiate the given function and find the critical points here both types of critical points are present, then we further input two critical points if they are in our given boundary conditions back into $f(x)$ and get our value of maxima and minima. We have to remember only one value can be maximum so the other one will give a minimum which we do not need for the question.

Complete step-by-step answer :
The given function is $f(x) = 2x - 3{x^{\dfrac{2}{3}}} + 2$ first we will try to integrate it with our standard formula for differentiation, so we will write :
$$ $f'(x) = 2 - 3 \times \dfrac{2}{3}{x^{ - \dfrac{1}{3}}}$
Which further gives:
$f'(x) = 2 - 3 \times \dfrac{2}{3}{x^{ - \dfrac{1}{3}}}$
$f'(x) = 2 - 2{x^{ - \dfrac{1}{3}}}$
We see that
$f'(x) = 0$when $x = 1$ and $f'(x)$ does not exist when $x = 0$ so we
Put these value of $x$ into the $f(x)$ we get
$f(0) = 2$ and $f(1) = 1$
So we get the maximum value of the given function $f(x) = 2x - 3{x^{\dfrac{2}{3}}} + 2$ as 2 on the value of $0$ . Remember here that the two points we found out which are $0,1$ are inside the interval given to us $[ - 1,3]$ . These two points are called critical points.
So, the correct answer is “ $0,1$ ”.

Note : Always keep in mind that during calculation of maxima or minima the critical points should always be inside the given interval. Any values outside this interval must be rejected.