Answer
$f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{3x}{4}+\frac{23}{15}$
Work Step by Step
$f’’(x)=x^3-2x,$ $f’(1)=0,$ $f(1)=2$
$f’(x)=\int(x^3-2x)dx$
$f’(x)=\frac{x^4}{4}-x^2+C_1$
Substitute in $f’(1)=0$ to solve for $C_1$
$0=\frac{1}{4}-1+C_1$
$C_1=\frac{3}{4}$
$f’(x)=\frac{x^4}{4}-x^2+\frac{3}{4}$
$f(x)=\int(\frac{x^4}{4}-x^2+\frac{3}{4})dx$
$f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{3x}{4}+C_2$
Substitute in $f(1)=2$ to solve for $C_2$
$2=\frac{1}{20}-\frac{1}{3}+\frac{3}{4}+C_2$
$C_2=\frac{23}{15}$
$f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{3x}{4}+\frac{23}{15}$
