## Calculus (3rd Edition)

$f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{x^2}{2}-\frac{x}{4}+\frac{121}{30}$
$f’’(x)=x^3-2x+1,$ $f’(1)=0,$ $f(1)=4$ $f’(x)=\int(x^3-2x+1)dx$ $f’(x)=\frac{x^4}{4}-x^2+x+C_1$ Substitute in $f’(1)=0$ to solve for $C_1$ $0=\frac{1}{4}-1+1+C_1$ $C_1=-\frac{1}{4}$ $f’(x)=\frac{x^4}{4}-x^2+x-\frac{1}{4}$ $f(x)=\int(\frac{x^4}{4}-x^2+x-\frac{1}{4})dx$ $f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{x^2}{2}-\frac{x}{4}+C_2$ Substitute in $f(1)=4$ to solve for $C_2$ $4=\frac{1}{20}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+C_2$ $C_2=\frac{121}{30}$ $f(x)=\frac{x^5}{20}-\frac{x^3}{3}+\frac{x^2}{2}-\frac{x}{4}+\frac{121}{30}$