Chapter 5.1, Problem 21E

To determine

To find: theextreme values of the function and where they occur.

Answer to Problem 21E

the local minimum value is $-\frac{41}{27}$ at $x=\frac{4}{3}$ and local maximum value is $17$ at $x=-2$ .

Explanation of Solution

Given information:

The function is $y=x^3+x^2-8x+5$

Calculation: the function is defined for each value of x, the domain of the function has no endpoints so all the extreme values must occur at critical points,

To find the extreme values of function, first find its derivative,

  $\begin{array}{l}y=x^3+x^2-8x+5\\ y\text{'}=3x^2-2x-8\text{ [differentiate and simplify]}\end{array}$

For critical points equal the derivative of function to zero,

  $\begin{array}{l}3x^2+2x-8=0\\ 3x^2+6x-4x-8=0\text{ [factor]}\\ 3x\left(x+2\right)-4\left(x-2\right)=0\text{ [simplify]}\\ x=-2,\frac{4}{3}\text{ [simplify]}\end{array}$

Now substitute in function,

  $\begin{array}{l}y=x^3+x^2-8x+5\\ y=-8+4+16+5\text{ [substitute }x=-2\text{ ]}\\ y=17\text{ [simplify]}\\ y={\left(\frac{4}{3}\right)}^3+{\left(\frac{4}{3}\right)}^2-8\left(\frac{4}{3}\right)+5\text{ [substitute }x=\frac{4}{3}\text{ ]}\\ y=-\frac{64}{27}+\frac{16}{9}+\frac{32}{3}+5\text{ [simplify]}\\ y=-\frac{41}{27}\end{array}$

Thus, the local minimum value is $-\frac{41}{27}$ at $x=\frac{4}{3}$ and local maximum value is $17$ at $x=-2$ .