Question

The mean value theorem applies to the given function on the given interval. How do you find all possible values of $f\left( x \right)={{x}^{\dfrac{9}{4}}}$ on the interval $\left[ 0,1 \right]$?

Answer 1

From the question given the mean value theorem is applicable to the function $f\left( x \right)={{x}^{\dfrac{9}{4}}}$ and on the given interval $\left[ 0,1 \right]$. By this we have to find the all-possible values of the function in the given interval. The mean value theorem states that if $f\left( x \right)$ is a continuous function on the interval $\left[ a,b \right]$ and differentiable on the interval $\left( a,b \right)$ then all the possible values are $c\in \left( a,b \right)$ such that ${{f}^{|}}\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$. By this theorem we can find all the possible values of the given function in the given interval.

Complete step-by-step answer:
From the question given function is
$\Rightarrow f\left( x \right)={{x}^{\dfrac{9}{4}}}$
The given interval is
$\left[ 0,1 \right]$
And also given that the function is applicable for the men value theorem that means the mean value theorem is applicable for the given function in the given interval.
As we know that the mean value theorem states that if $f\left( x \right)$ is a continuous function on the interval $\left[ a,b \right]$ and differentiable on the interval $\left( a,b \right)$ then all the possible values are $c\in \left( a,b \right)$ such that
$\Rightarrow {{f}^{|}}\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$
Here the function is
$\Rightarrow f\left( x \right)={{x}^{\dfrac{9}{4}}}$
And the values of a, b are
$\Rightarrow a=0$
$\Rightarrow b=1$
The function is continuous on the interval $\left[ 0,1 \right]$ and differentiable on the interval $\left( 0,1 \right)$
By this we can say that
$\Rightarrow f\left( 0 \right)=0$
$\Rightarrow f\left( 1 \right)=1$
The differentiation of the given function
$\Rightarrow {{f}^{|}}\left( x \right)=\dfrac{9}{4}{{x}^{\dfrac{9}{4}-1}}=\dfrac{9}{4}{{x}^{\dfrac{5}{4}}}$
Therefore, according to the, mean value theorem
$\Rightarrow {{f}^{|}}\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$
$\Rightarrow \dfrac{9}{4}{{c}^{\dfrac{5}{4}}}=\dfrac{f\left( 1 \right)-f\left( 0 \right)}{1-0}=\dfrac{1-0}{1}=1$
$\Rightarrow {{c}^{\dfrac{5}{4}}}=\dfrac{4}{9}$
Taking the natural logs on both sides
$\Rightarrow \dfrac{5}{4}\ln c=\ln \left( \dfrac{4}{9} \right)$
$\Rightarrow \ln c=\dfrac{4}{5}\ln \left( \dfrac{4}{9} \right)$
$\Rightarrow \ln c=-0.81$
$\Rightarrow c=0.44$
Therefore, the required answer is $c=0.44$ and it belongs to $\left( 0,1 \right)$

Note: Students should know the basics theorems like mean value theorem and students should remember the conditions of the mean value theorem if in case any function is continuous in the interval [a,b] and not differentiable in the interval (a,b) then the mean value theorem can not be implemented.