What are the local extrema of f(x)= -x^3 + 3x^2 + 10x + 13?

1 Answer
Nov 4, 2015

Local maximum is $25 + \frac{26 \sqrt{\frac{13}{3}}}{3}$
Local minimum is $25 - \frac{26 \sqrt{\frac{13}{3}}}{3}$

Explanation:

To find local extrema, we can use the first derivative test. We know that at a local extrema, at the very least the function's first derivative will equal zero. So, let's take the first derivative and set it equal to 0 and solve for x.

$f \left(x\right) = - {x}^{3} + 3 {x}^{2} + 10 x + 13$

$f ' \left(x\right) = - 3 {x}^{2} + 6 x + 10$

$0 = - 3 {x}^{2} + 6 x + 10$

This equality can be solved easily with the quadratic formula. In our case, $a = - 3$, $b = 6$ and $c = 10$

Quadratic formula states:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

If we plug back our values into the quadratic formula, we get
$x = \frac{- 6 \pm \sqrt{156}}{-} 6 = 1 \pm \frac{\sqrt{156}}{6} = 1 \pm \sqrt{\frac{13}{3}}$

Now that we have the x values of where the local extrema are, let's plug them back into our original equation to get:

$f \left(1 + \sqrt{\frac{13}{3}}\right) = 25 + \frac{26 \sqrt{\frac{13}{3}}}{3}$ and

$f \left(1 - \sqrt{\frac{13}{3}}\right) = 25 - \frac{26 \sqrt{\frac{13}{3}}}{3}$