Question

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

Answer 1

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

Explanation:

`f'(x) = (x-8)^9 (x-4)^7 (x+3)^7` so the stationary points are

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

`f(x) approx f(x_0) + C (x-x_0)^n` where `n` is defined by the firts non null derivative of `f(x)` 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 = 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 = 159638904832/5 > 0` minimum

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