Question

How do you determine if Rolle’s Theorem applies to the given function ${{x}^{3}}-9x$ on $\left[ 0,3 \right].$ If so, how do you find all numbers $c$ on the interval that satisfy the theorem?

Answer 1

Hint: We check if the given function satisfies the hypotheses of the theorem. We verify that the given function is continuous on the given interval, closed at both ends. We check if the given function is differentiable on the given interval, open at both ends. Also, we have to verify if the function values at both end points are the same. If these conditions are satisfied, then comes the conclusion.

Complete step by step answer:
Consider the given function ${{x}^{3}}-9x.$
Suppose that $f\left( x \right)={{x}^{3}}-9x.$
Now we have to verify all the hypotheses of Rolle’s Theorem.
First hypothesis of Rolle’s theorem says about the continuity of the function on the given closed interval.
Here the given interval is the closed interval $\left[ 0,3 \right].$
Now we have to verify if the function $f\left( x \right)={{x}^{3}}-9x$ is continuous on the closed interval $\left[ 0,3 \right].$
The given function $f\left( x \right)={{x}^{3}}-9x$ is continuous on the interval $\left[ 0,3 \right].$ Because, $f\left( x \right)={{x}^{3}}-9x$ is a polynomial. [Every polynomial is continuous at every real number.]
The first hypothesis is satisfied.
Now, we have to verify the second hypothesis of Rolle’s Theorem that says about the differentiability of the function on the given interval with both ends open.
That is, we have to check if the given function $f\left( x \right)={{x}^{3}}-9x$ is differentiable on the open interval $\left( 0,3 \right).$
Therefore, we differentiate the given function as follows
$\Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( {{x}^{3}}-9x \right).$
When we use the linearity property $\dfrac{d\left( ax+y \right)}{dx}=a\dfrac{dx}{dx}+\dfrac{dy}{dx},$ this will become,
$\Rightarrow {f}'\left( x \right)=\dfrac{d{{x}^{3}}}{dx}-9\dfrac{dx}{dx}.$
We use the rules, $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ and $\dfrac{dx}{dx}=1,$ then,
$\Rightarrow {f}'\left( x \right)=3{{x}^{2}}-9.$
This implies that the derivative exists for the given function $f\left( x \right)={{x}^{3}}-9x.$
Therefore, the function $f\left( x \right)={{x}^{3}}-9x$ is differentiable on $\left( 0,3 \right).$
Thus, the second hypothesis is satisfied.
Now we have to check if the function values at the end points are the same.
The left end point is $0$ and the right end point is $3.$
We find $f\left( 0 \right)$ and $f\left( 3 \right).$
$\Rightarrow f\left( 0 \right)={{0}^{3}}-9\times 0=0$
And.
$\Rightarrow f\left( 3 \right)={{3}^{3}}-9\times 3=27-27=0$
Therefore, $f\left( 0 \right)=f\left( 3 \right).$
That means, all the hypotheses are satisfied.
Let us check if the function satisfies the conclusion of the theorem.
It says that there exists at least one $c$ in the given open interval that satisfies ${f}'\left( c \right)=0.$
Suppose that there exists a point $c$ whose first derivative is $0.$
Let us find the derivative of the function at the point $c.$
That is,
$\Rightarrow {f}'\left( c \right)=3{{c}^{2}}-9.$
We equate this to $0.$
$\Rightarrow {f}'\left( c \right)=3{{c}^{2}}-9=0.$
That will give us,
$\Rightarrow 3{{c}^{2}}=9.$
And we get,
$\Rightarrow {{c}^{2}}=\dfrac{9}{3}=3.$
So,
$\Rightarrow c=\pm \sqrt{3}.$
Therefore, the possible values of $c$ are $-\sqrt{3}$ and $\sqrt{3}.$
Since $-\sqrt{3}\notin \left( 0,3 \right),$ the value of $c=\sqrt{3}.$
Hence, there exists a $c=\sqrt{3}$ in $\left( 0,3 \right)$ that satisfies ${f}'\left( c \right)=0.$
Hence, Rolle’s Theorem applies to the given function ${{x}^{3}}-9x.$

Note:
Rolle’s Theorem: Suppose that a function $f\left( x \right)$ is continuous on the closed interval $\left[ a,b \right]$ and differentiable on the open interval $\left( a,b \right).$ Then there exists at least one point $c$ in the open interval $\left( a,b \right)$ such that ${f}'\left( c \right)=0.$