Answer
Verified
410.7k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE