# A differentiable function f has only one critical number: x=-3. Identify the relative extrema of f at (-3, f(-3)) if f'(-4)=(1/2) and f'(-2)=-1?

Apr 20, 2017

$f \left(- 3\right)$ is a relative maximum.

#### Explanation:

$f ' \left(- 3\right)$ is either undefined or $0$.

Since $f$ is differentiable, $f '$ is never undefined. That, together with "$- 3$ is a critical number", implies that $f ' \left(- 3\right) = 0$.

Derivative have the intermediate value property.

Therefore, since $f ' \left(- 4\right) > 0$ , there is nowhere in $\left[- 4 , 0\right)$ where $f '$ is $0$, we can conclude that $f ' \left(x\right) > 0$ for $x \in$[-4,-3)$s o$f$i s \in c r e a \sin g o n$[-4,-3).

Similarly, since $f ' \left(- 2\right) < 0$ , we can conclude that $f ' \left(x\right) < 0$ for $x \in$(-3,-2]$s o$f$i s \mathrm{de} c r e a \sin g o n$(-3,-2].

Therefore $f \left(- 3\right)$ is a relative maximum.