Question

determine whether rolle’s theorem applies to the function. if it applies find all possible values of c

Answer 1

Answer:

• Yes

,

• c=0

Explanation:

Given the function, G(x):

$G(x)=\frac{4}{x^2+16}$

Rolle's Theorem:

According to this theorem if the given function is:

• continuous in [a,b]

,

• differentiable in (a,b)

,

• f(a) = f(b)

then, there exist c in (a,b) such that f'(c)=0.

Continuity of function:

Since the given function is continuous function, it is continuous everywhere. Therefore, G(x) is continuous in [-2,2]

Differentiability

The rational function is differentiable using the quotient rule. Therefore, G(x) is differentiable in (-2,2).

Next, evaluate G(-2) and G(2):

$\begin{gathered} G(-2)=\frac{4}{(-2)^2+16}=\frac{4}{4+16}=\frac{4}{20}=\frac{1}{5} \\ G(2)=\frac{4}{(2)^2+16}=\frac{4}{4+16}=\frac{4}{20}=\frac{1}{5} \\ G(-2)=G(2) \end{gathered}$

Thus, Rolle's theorem applies on G(x).

Next, we find the possible values of c.

By Rolle's theorem, there exist c in (a,b) such that f'(c) = 0.

$\begin{gathered} G(x)=\frac{4}{x^2+16} \\ G(x)=4(x^2+16)^{-1} \\ \text{Let u=}x^2+16\implies G(u)=4u^{-1} \\ \frac{dG}{dx}=\frac{dG}{du}\times\frac{du}{dx}=-4u^{-2}\times2x=-8x(x^2+16)^{-2} \\ G^{\prime}(x)=-\frac{8x}{(x+16)^2} \end{gathered}$

Thus:

$\begin{gathered} G^{\prime}(c)=-\frac{8c}{(c+16)^2}=0 \\ \implies-8c=0 \\ \implies c=0 \end{gathered}$

Since c should be in the interval [-2,2], the value of c that satisfies the theorem is 0.