`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.
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.