# `f(x) = xsqrt(x + 4)` 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 7 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the two x intercepts of the function, hence, you need to solve for x the equation f(x) = 0, such that:

`f(x) = x*sqrt(x+4) = 0`

`x*sqrt(x+4) = 0 x = 0`

`sqrt(x+4) = 0 => x + 4 = 0^2 => x + 4 = 0 => x = -4`

You need to evaluate the derivative of the function, using the product rule:

`f'(x) = x'*sqrt(x+4) +  x*(sqrt(x+4))' => f'(x) = sqrt(x+4) + x/(2sqrt(x+4))`

You need to solve for x the equation f'(x) = 0:

` sqrt(x+4) + x/(2sqrt(x+4)) = 0 => (2(x+4) + x)/(2sqrt(x+4)) = 0`

`2x + 8 + x = 0 => 3x = -8 => x = -8/3 = -2.(6)`

Notice that -2.(6) is found between x intercepts -4 and 0.

Hence, the derivative of the function cancels at x = -2.(6), which is found between the x intercepts -4 and 0.