# How do you find local maximum value of f using the first and second derivative tests: f(x)= x^2 + 8x -12?

Dec 20, 2015

$f$ has no local maxima. It has a local minimum when $x = - 4$.

#### Explanation:

Find $f ' \left(x\right)$ and see when it $= 0$ or $\text{DNE}$. These are the critical numbers at which a local maximum could exist.

To determine whether or not the point is a local maximum, you could use either the first or second derivative tests.

For the first derivative test , create a sign chart where the important values are the critical numbers. If the signs change from positive to negative, the point is a local maximum.

For the second derivative test , plug the critical number(s) into the second derivative. If the value is negative, the function is concave down at that point meaning the point is a local maximum.

Now, let's do the work:

$f ' \left(x\right) = 2 x + 8$

$f ' \left(x\right) = 0$ when $x = - 4$.
$f ' \left(x\right)$ never $\text{DNE}$.

First derivative test:

$\textcolor{w h i t e}{\times \times x X X X \times} - 4$
$\leftarrow - - - - - - - - - - \rightarrow$
$\textcolor{w h i t e}{\times x} \text{NEGATIVE"color(white)(xxxxx)"POSITIVE}$

Since the sign of the first derivative goes from negative to positive, there is a local minimum when $x = - 4$.

We can prove the same thing with the...

Second derivative test:

$f ' ' \left(x\right) = 2$

Thus, the second derivative is ALWAYS positive, and the function is always concave up, which results in a local minimum.

Therefore, the function has no local maxima.

We can check a graph, even though it is obvious that the graph will form a parabola facing up, which will have only a minimum at its vertex:
graph{x^2+8x-12 [-65, 66.67, -36.6, 29.23]}