Question

How do you find the critical numbers of $f\left( x \right) = {x^{\dfrac{2}{3}}} + {x^{ - \dfrac{1}{3}}}$?

Answer 1

Hint:In order to determine the critical numbers for the above function, first find the derivative of the function with respect to x . Put the derivative equal to zero to find out the value of $x$. The values of $x$ are nothing but the critical number of $f\left( x \right)$
Formula:
$\dfrac{d}{{dx}}(\ln x) = \dfrac{1}{x}$
$\dfrac{d}{{dx}}({e^x}) = {e^x}$
$\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$

Complete step by step solution:
We are given a function $f\left( x \right) = {x^{\dfrac{2}{3}}} + {x^{ - \dfrac{1}{3}}}$
In order to find the critical number of the above function, we first know what are critical numbers.
Critical numbers of any function $f\left( x \right)$ are the values of variable x for which derivative of
$f'(x) = 0$.
For this, we have to first find out the derivative of our function with respect to .
$\dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {{x^{\dfrac{2}{3}}} + {x^{ - \dfrac{1}{3}}}}
\right)$
Separating the derivative inside the bracket , we get
$f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^{\dfrac{2}{3}}}} \right) + \dfrac{d}{{dx}}\left( {{x^{ - \dfrac{1}{3}}}} \right)$
As we know the derivative of variable $x$raised to power some value \[n\] is $\dfrac{d}{{dx}}({x^n})
= n{x^{n - 1}}$. Applying this rule to the above equation to find the derivative of both the terms, we get
$
f'\left( x \right) = \dfrac{2}{3}{x^{\dfrac{2}{3} - 1}} + \dfrac{1}{3}{x^{ - \dfrac{1}{3} - 1}} \\
= \dfrac{2}{3}{x^{\dfrac{{2 - 3}}{3}}} + \dfrac{1}{3}{x^{ - \dfrac{{1 - 3}}{3}}} \\
= \dfrac{2}{3}{x^{\dfrac{{ - 1}}{3}}} + \dfrac{1}{3}{x^{\dfrac{{ - 4}}{3}}} \\
$
Now putting the $f'(x) = 0$ to obtain the critical numbers
$
f'(x) = \dfrac{2}{3}{x^{\dfrac{{ - 1}}{3}}} + \dfrac{1}{3}{x^{\dfrac{{ - 4}}{3}}} = 0 \\
\dfrac{2}{3}{x^{\dfrac{{ - 1}}{3}}} + \dfrac{1}{3}{x^{\dfrac{{ - 4}}{3}}} = 0 \\
$
Multiplying both sides of the equation with $\dfrac{3}{{{x^{ - \dfrac{1}{3}}}}}$, our equation
becomes
$\left( {\dfrac{3}{{{x^{ - \dfrac{1}{3}}}}}} \right)\left( {\dfrac{2}{3}{x^{\dfrac{{ - 1}}{3}}} +
\dfrac{1}{3}{x^{\dfrac{{ - 4}}{3}}}} \right) = 0 \times \dfrac{3}{{{x^{ - \dfrac{1}{3}}}}}$
Simplifying further by using the rule of exponent that $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
$
\left( {\dfrac{3}{{{x^{ - \dfrac{1}{3}}}}}} \right)\left( {\dfrac{2}{3}{x^{\dfrac{{ - 1}}{3}}} +
\dfrac{1}{3}{x^{\dfrac{{ - 4}}{3}}}} \right) = 0 \times \dfrac{3}{{{x^{ - \dfrac{1}{3}}}}} \\
2\left( {\dfrac{{{x^{\dfrac{{ - 1}}{3}}}}}{{{x^{ - \dfrac{1}{3}}}}}} \right) + \dfrac{{{x^{\dfrac{{ -
4}}{3}}}}}{{{x^{ - \dfrac{1}{3}}}}} = 0 \\
2\left( {{x^{\dfrac{{ - 1}}{3} + \dfrac{1}{3}}}} \right) + {x^{\dfrac{{ - 4}}{3} + \dfrac{1}{3}}} = 0
\\
2\left( {{x^0}} \right) + {x^{\dfrac{{ - 3}}{3}}} = 0 \\
$
As we know anything raised to the power zero equal to one
$
2 + {x^{ - 1}} = 0 \\
{x^{ - 1}} = - 2 \\
\dfrac{1}{x} = - 2 \\
$
Taking reciprocal on both of the sides, we get
$x = - \dfrac{1}{2}$
Therefore, the critical number for function$f\left( x \right) = {x^{\dfrac{2}{3}}} + {x^{ -
\dfrac{1}{3}}}$is $x = - \dfrac{1}{2}$.
Additional Information:
1.What is Differentiation?
It is a method by which we can find the derivative of the function .It is a process through which we can find the instantaneous rate of change in a function based on one of its variables. Let y = f(x) be a function of x. So the rate of change of $y$per unit change in $x$ is given by:
$\dfrac{{dy}}{{dx}}$.

Note:
1.Don’t forget to cross-check your answer at least once.
2.Differentiation is basically the inverse of integration.
3. Critical numbers are those values of x at which the graph of function changes.