Answer
local maximum: $f(x)=17$ and $x=-2$
local minimum: $f(x)=\frac{-41}{27}$ and $x=\frac{4}{3}$
Work Step by Step
Given $y=x^3+x^2-8x+5$
the critical point is where the first derivative is zero:
so $\frac{dy}{dx}=\frac{d{(x^3+x^2-8x+5)}}{dx}$
${\frac{dy}{dx}}=3x^2+2x-8$
so the critical point is: ${\frac{dy}{dx}}=0$
$f'(x)=3x^2+2x-8=0$
$x=-2,x=\frac{4}{3}$
the critical value of y can be obtained by putting in the value of x:
$y=(-2)^3+(-2)^2-8\times(-2)+5=17$
$y=(\frac{4}{3})^3+(\frac{4}{3})^2-8\times(\frac{4}{3})+5=\frac{-41}{27}$
Thus we have:
local maximum: $f(x)=17$ and $x=-2$
local minimum: $f(x)=\frac{-41}{27}$ and $x=\frac{4}{3}$
