I guess the function is [tex]f(x)=9-x^{2/3}[/tex]. Then [tex]f(-27)=0[/tex] and [tex]f(27)=0[/tex].
The derivative is [tex]f'(x)=-\dfrac23 x^{-1/3}[/tex], but there is no [tex]c[/tex] such that
[tex]-\dfrac23c^{-1/3}=0[/tex]
This doesn't contradict Rolle's theorem because [tex]f'(0)[/tex] does not exists. In other words, [tex]f[/tex] is not differentiable on (-27, 27), so the conditions of Rolle's theorem are not met. (Looks like that would be the last option, or the second to last option if the last one is "Nothing can be concluded")
Rolle's theorem are used to determine if a function is differentiable and at the same time continuous, or not.
The conclusion that can be made is:
(d) This does not contradict Rolle's Theorem, since f '(0) does not exist, and so f is not differentiable on (−27, 27).
The function is given as:
[tex]\mathbf{f(x) = 9 - x^\frac{2}3}[/tex]
Start by calculating f(-27) and f(27)
[tex]\mathbf{f(-27) = 9 - (-27)^\frac23 = 9 - 9 = 0}[/tex]
[tex]\mathbf{f(27) = 9 - (-27)^\frac23 = 9 - 9 = 0}[/tex]
Next, differentiate [tex]\mathbf{f(x) = 9 - x^\frac{2}3}[/tex]
[tex]\mathbf{f'(x) = - \frac23x^{\frac23 - 1}}[/tex]
[tex]\mathbf{f'(x) = - \frac23x^{-\frac13 }}[/tex]
Substitute c for x
[tex]\mathbf{f'(c) = - \frac23c^{-\frac13 }}[/tex]
Set the derivative to 0
[tex]\mathbf{ - \frac23c^{-\frac13 } = 0}[/tex]
Divide both sides by -2/3
[tex]\mathbf{ c^{-\frac13 } = 0}[/tex]
Take -3rd root of both sides
[tex]\mathbf{ c = 0^{-3}}[/tex]
[tex]\mathbf{ c =DNE}[/tex] ---- does not exist
So, we have the following observations:
[tex]\mathbf{ f(-27) = f(27)}[/tex]
[tex]\mathbf{f'(c) = 0}[/tex] ------ does not exist
[tex]\mathbf{c \ne (-27,27))}[/tex]
From the list of given options, the option that supports the above conclusions is: (d)
Read more about Rolle's theorem at:
https://brainly.com/question/4853582
