Chapter 12.2, Problem 61E

To determine

For the function $f\left(x\right)=\frac{1}{6}{x}^3-\frac{7}{2}{x}^2+\frac{45}{2}x-12$; $3\le x\le 15$ here $x=3$ corresponds to the year 2003.

The first derivative of $f\left(x\right)=\frac{1}{6}{x}^3-\frac{7}{2}{x}^2+\frac{45}{2}x-12$.

The critical numbers of $f\left(x\right)=\frac{1}{6}{x}^3-\frac{7}{2}{x}^2+\frac{45}{2}x-12$.

The second derivative of $f\left(x\right)=\frac{1}{6}{x}^3-\frac{7}{2}{x}^2+\frac{45}{2}x-12$.

The local minimum and maximum of $f\left(x\right)=\frac{1}{6}{x}^3-\frac{7}{2}{x}^2+\frac{45}{2}x-12$.