# How do use the first derivative test to determine the local extrema f(x) = (x+1)(x-3)^2?

Aug 31, 2015

Have a look:

#### Explanation:

You can use the Product and Chain Rule to find the derivative:
$f ' \left(x\right) = 1 \cdot {\left(x - 3\right)}^{2} + \left(x + 1\right) \cdot 2 \left(x - 3\right) = {x}^{2} - 6 x + 9 + 2 \left({x}^{2} - 3 x + x - 3\right) =$
$= {x}^{2} - 6 x + 9 + 2 \left({x}^{2} - 2 x - 3\right) = {x}^{2} - 6 x + 9 + 2 {x}^{2} - 4 x - 6 =$
$= 3 {x}^{2} - 10 x + 3$
Now you can set it equal to zero to find the x coordinate(s) of the local extrema:
$3 {x}^{2} - 10 x + 3 = 0$
We can use the Quadratic Formula:
${x}_{1 , 2} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm 8}{6} =$
So you get:
${x}_{1} = 3$ giving $y = 0$
${x}_{2} = \frac{1}{3}$ giving $y = 9.5$

Graphically:
graph{(x+1)((x-3)^2) [-10, 10, -5, 5]}