# How do use the first derivative test to determine the local extrema f(x) = (x+3)(x-4)^2?

Sep 25, 2015

See the explanation.

#### Explanation:

$f \left(x\right) = \left(x + 3\right) {\left(x - 4\right)}^{2}$

$f ' \left(x\right) = \left[1\right] {\left(x - 4\right)}^{2} + \left(x + 3\right) \left[2 \left(x - 4\right) \left(1\right)\right]$

$= \left(x - 4\right) \left[\left(x - 4\right) + \left(x + 3\right) 2\right]$

$= \left(x - 4\right) \left(3 x + 2\right)$

The critical numbers for $f$ are $4 \text{ and } - \frac{2}{3}$

We look at the sign of $f '$ on each interval to determine whether $f$ is increasing or decreasing on the interval

{: (bb "Interval", bb"Sign of "f',bb" Incr/Decr"), ((-oo,-2/3)," " +" ", " "" Incr"), ((-3/2,4), " " -, " " " Decr"), ((4,oo), " " +, " "" Incr") :}

$f$ has a local maximum of $\frac{1372}{27}$ at $- \frac{2}{3}$

and a local minimum of $0$ at $4$.