## University Calculus: Early Transcendentals (3rd Edition)

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 $x=\sqrt {\frac{2}{3}}$
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 $x=\sqrt {\frac{2}{3}}$