Answer
local maximum $f(x)=4+4\frac{\sqrt{6}}{9}$ and $x=-\sqrt {\frac{2}{3}}$
local minimum $f(x)=4-4\frac{\sqrt{6}}{9}$ and $x=\sqrt {\frac{2}{3}}$
Work Step by Step
Given $y=x^3-2x+4$
The critical point is where the first derivative is zero:
$\frac{dy}{dx}=\frac{d{(x^3-2x+4)}}{dx}$
${\frac{dy}{dx}}=3x^2-2$
so the critical point is ${\frac{dy}{dx}}=0$
$f'(x)=3x^2-2=0$
$x=\sqrt {\frac{2}{3}},-\sqrt {\frac{2}{3}}$
the critical value of y can be obtained by putting in the value of x:
$y=x^3-2x+4=(\sqrt {\frac{2}{3}})^3-2(\sqrt {\frac{2}{3}})+4=4-4\frac{\sqrt{6}}{9}$
$y=x^3-2x+4=(-\sqrt {\frac{2}{3}})^3-2(-\sqrt {\frac{2}{3}})+4=4+4\frac{\sqrt{6}}{9}$
thus we have: local maximum $f(x)=4+4\frac{\sqrt{6}}{9}$ and $x=-\sqrt {\frac{2}{3}}$
local minimum $f(x)=4-4\frac{\sqrt{6}}{9}$ and $x=\sqrt {\frac{2}{3}}$
