How do use the first derivative test to determine the local extrema f(x)= x^3 - x^2 - 40x + 8?

Sep 13, 2015

This function has a local maximum value of $\frac{2516}{27}$ at $x = - \frac{10}{3}$ and a local minimum value of $- 104$ at $x = 4$.

Explanation:

The first derivative is $f ' \left(x\right) = 3 {x}^{2} - 2 x - 40$. This can be factored as $f ' \left(x\right) = 3 {x}^{2} - 2 x - 40 = \left(3 x + 10\right) \left(x - 4\right)$. Setting this equal to zero and solving for $x$ gives two critical points for $f$ at $x = - \frac{10}{3}$ and $x = 4$.

You can check that $f '$ changes sign from positive to negative as $x$ increases through $x = - \frac{10}{3}$ and negative to positive as $x$ increases through $x = 4$ (note that $f '$ is a continuous function and that, for instance, $f ' \left(- 4\right) = 48 + 8 - 40 = 16 > 0$, $f ' \left(0\right) = - 40 < 0$, and $f ' \left(5\right) = 75 - 10 - 40 = 25 > 0$).

Therefore, the First Derivative Test implies that $f$ has a local maximum at $x = - \frac{10}{3}$ and a local minimum at $x = 4$.

The local maximum value (output) is $f \left(- \frac{10}{3}\right) = - \frac{1000}{27} - \frac{100}{9} + \frac{400}{3} + 8 = \frac{2516}{27}$ and the local minimum value (output) is $f \left(4\right) = 64 - 16 - 160 + 8 = - 104$.