# If f'(x) = (x-8)^9 (x-4)^7 (x+3)^7, what are the local minima and maxima of f(x)?

Aug 19, 2016

There is a local maximum at $x = 4$ and two local minima at
$x = - 3$ and $x = 8$

#### Explanation:

$f ' \left(x\right) = {\left(x - 8\right)}^{9} {\left(x - 4\right)}^{7} {\left(x + 3\right)}^{7}$ so the stationary points are

$x = - 3 , x = 4$ and $x = 8$. At those points the behavior of $f \left(x\right)$ is described by

$f \left(x\right) \approx f \left({x}_{0}\right) + C {\left(x - {x}_{0}\right)}^{n}$ where $n$ is defined by the firts non null derivative of $f \left(x\right)$ at ${x}_{0}$, and $C$ is 1/(n!)(d^nf(x_0))/(dx^n). In the present case we have:

1) For the stationary point $x = - 3$ we have $n = 8$ and
$C = \frac{1941871315289213}{8} > 0$ minimum

2) For the stationary point $x = 4$ we have $n = 8$ and
$C = - 26985857024 < 0$ maximum

3) For the stationary point $x = 8$ we have $n = 10$ and
$C = \frac{159638904832}{5} > 0$ minimum

So there is a local maximum at $x = 4$ and two local minima at
$x = - 3$ and $x = 8$