Question

find the maximum and minimums and identify as absolute or local. if there arent any, write none.

1. $f(x)=4 - x + x^{2}$

$f(x)=x^{2}-x + 4$
$f(x)=2x - 1 = 0$
$2x = 1$
$x=\frac{1}{2}$
$f(x)=2>0$. max:
min: local min

1. $g(x)=(x + 3)(x - 1)^{2}$
2. max:

min:

Explanation:

Step1: Expand $g(x)$

$g(x)=(x + 3)(x - 1)^2=(x + 3)(x^{2}-2x + 1)=x^{3}-2x^{2}+x+3x^{2}-6x + 3=x^{3}+x^{2}-5x + 3$

Step2: Find the derivative

$g'(x)=3x^{2}+2x - 5=(3x + 5)(x - 1)$

Step3: Set derivative to 0

$3x+5 = 0$ gives $x=-\frac{5}{3}$; $x - 1=0$ gives $x = 1$

Step4: Find second - derivative

$g''(x)=6x+2$

Step5: Evaluate second - derivative at critical points

$g''(-\frac{5}{3})=6\times(-\frac{5}{3})+2=-10 + 2=-8<0$, so local max at $x = -\frac{5}{3}$
$g''(1)=6\times1+2=8>0$, so local min at $x = 1$

Answer:

1. Max: Local max at $x=-\frac{5}{3}$

Min: Local min at $x = 1$