`f(x) = -3xsqrt(x + 1)` Find the two x-intercepts of the function `f` and show that `f'(x) = 0` at some point between the two x-intercepts.

Textbook Question

Chapter 3, 3.2 - Problem 8 - Calculus of a Single Variable (10th Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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Hello!

f(x) = 0 for x=0 and x=-1 (obvious).

f is continuous on [-1, 0] and is differentiable on (-1, 0) (as an elementary function). So we can aplly Rolle's Theorem and conclude that there is at least one point in (-1, 0) where f'(x)=0.

Actually, we can find this point:

`f'(x) = -3sqrt(x+1) - 3x/(2sqrt(x+1)) = (-3/2)*(2(x+1) + x)/sqrt(x+1) = (-3/2)*(3x+2)/sqrt(x+1),`

therefore f'(-2/3)=0.

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