Question

The greatest value of $f(x) = \sqrt[3]{{(x + 1)}} - \sqrt[3]{{(x - 1)}}$ on $\left[ {0,{\text{ }}1} \right]$ is

Answer 1

We have to find the greatest value of $f\left( x \right)$ on $\left[ {0{\text{ }},{\text{ }}1} \right]$ . We solve this using the concept of applications of derivatives. To find the maximum value we first differentiate $f\left( x \right)$ with respect of and the equate the value of, for obtaining the greatest value of the function on $\left[ {{\text{ }}0{\text{ }},{\text{ }}1{\text{ }}} \right]{\text{ }}$
We will neglect the value of $x$ which does not lie in between the interval $\left[ {0{\text{ }},{\text{ }}1} \right]$ .

Complete step-by-step solution:
Given :
$f(x) = \sqrt[3]{{(x + 1)}} - \sqrt[3]{{(x - 1)}}$
For the greatest value of$f\left( x \right)$, we differentiate $f\left( x \right)$ with respect to.
Now , differentiating$f\left( x \right)$ with respect to
Using the formula $\dfrac{d}{{dx}}{x^n} = n{x^{(n - 1)}}$
Using this formula and applying it in$f\left( x \right)$, we get
$f'(x) = (\dfrac{1}{3}) \times {(x + 1)^{(\dfrac{{ - 2}}{3})}} - (\dfrac{1}{3}) \times {(x - 1)^{(\dfrac{{ - 2}}{3})}}$
Now , for the point of greatest value
Put $f’(x)=0$
So, $(\dfrac{1}{3}) \times {(x + 1)^{(\dfrac{{ - 2}}{3})}} - (\dfrac{1}{3}) \times {(x - 1)^{(\dfrac{{ - 2}}{3})}} = 0$
On simplification , we get
$\dfrac{{[{{(x - 1)}^{(\dfrac{2}{3})}} - {{(x + 1)}^{(\dfrac{2}{3})}}]}}{{[3 \times {{({x^2} - 1)}^{(\dfrac{2}{3})}}]}} = 0$
The function does not exist for ${x^2} - 1 = 0$
does not exist at $x{\text{ }} = {\text{ }} \pm {\text{ }}1$
Now ,
$[{(x - 1)^{(\dfrac{2}{3})}} - {(x + 1)^{(\dfrac{2}{3})}}] = 0$
On further solving ,
${(x - 1)^{(\dfrac{2}{3})}} = {(x + 1)^{(\dfrac{2}{3})}}$
Cancelling the powers , we get
$\left( {{\text{ }}x{\text{ }} - {\text{ }}1{\text{ }}} \right){\text{ }} = {\text{ }}\left( {{\text{ }}x{\text{ }} + {\text{ }}1{\text{ }}} \right)$
The term of $x$ gets cancelled in the simplified expression , so there is no value of $x$ which lies on $[0{\text{ }},{\text{ }}1]$ .
So , the only values which can give the greatest value in the interval are $0$ and $1$ as they are included in the interval $[0{\text{ }},{\text{ }}1]$ .
But also , does not exist for $x{\text{ }} = {\text{ }}1$ ( as shown above )
So , the only possible value of $x$ is $x{\text{ }} = {\text{ }}0$
The value of function $f\left( x \right)$ at $x{\text{ }} = {\text{ }}0$ :
$f(0) = \sqrt[3]{{(1 - 0)}} - \sqrt[3]{{(0 - 1)}}$
$f\left( 0 \right){\text{ }} = {\text{ }}2$
Thus , the greatest value of the function $f\left( x \right)$ is $2$ .

Note: Using the property of increasing and decreasing function function we can compute that for what value of $x$ the function is decreasing and for what value of $x$ the function is increasing . If the first derivative of a function is positive for a value of $x$ then the particular value of $x$ gives the minimum value of the function and vice versa .