`f(x) = x^2 + 6x` Find the two x-intercepts of the function `f` and show that `f'(x) = 0` at some point between the two x-intercepts.

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Chapter 3, 3.2 - Problem 6 - 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^2 + 6x = 0`

You need to factor out x, such that:

`x(x + 6) = 0 => x = 0`

`x + 6 = 0 => x = -6`

You need to evaluate the derivative of the function:

`f'(x) = (x^2 + 6x)' => f'(x) = 2x + 6`

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

`2x + 6 = 0 => 2x = -6 => x = -3`

Notice that -3 is found between x intercepts -6 and 0.

Hence, the derivative of the function cancels at x = -3, which is found between the x intercepts -6 and 0.

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