Download Photo math it will help u a lot with steps
Answer:
(E)Nothing can be concluded.
Step-by-step explanation:
Given the function 

![f'(x)=-\dfrac{2}{3}x^{-\frac{1}{3}}\\f'(x)=-\dfrac{2}{3\sqrt[3]{x} }](https://tex.z-dn.net/?f=f%27%28x%29%3D-%5Cdfrac%7B2%7D%7B3%7Dx%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7D%5C%5Cf%27%28x%29%3D-%5Cdfrac%7B2%7D%7B3%5Csqrt%5B3%5D%7Bx%7D%20%7D)
If the derivative is set equal to zero, the function is undefined.
Nothing can be concluded since
and no such c in (-1,1) exists such that 
<u>THEOREM</u>
Rolle's theorem states that any real-valued differentiable function that attains equal values at two distinct points must have at least one stationary point somewhere between them—that is, a point where the first derivative is zero.
Answer:
multiplicative inverse
Step-by-step explanation:
To disembark means to leave / get off something (typically used when talking about vehicles)
e.g.:
The passengers disembarked the ship at 6:00.
As the fellow above said, the particle does make a U-turn at the vertex, and we can find that out by getting the derivative of s(t) and zeroing it out to get the point of the horizontal tangent.
Bearing in mind that ds/dt is really v(t) or the velocity equation, so when we zero out the ds/dt, is the same as saying the velocity went to 0, since it got zeroed out, and then the particle goes over the vertex and changes direction.

the negative acceleration value, simply means a decreasing rate of change, so, it means the particle is slowing down and possibly coming to a stop before changing direction again.