Chapter 7, Problem 36CLR

To determine

Tofind: the domain, range and zeros of the function.

Answer to Problem 36CLR

The domain, range and zeros of the function is $R,[-7,\infty )\text{ and }4\pm \sqrt{7}$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=x^2-8x+9$ .

Calculation: the function is defined for all real values of $x$ .

Thus, the domain of the function is $R$ .

To find the range, first differentiate the function,

  $\begin{array}{l}f\left(x\right)=x^2-8x+9\\ f\text{'}\left(x\right)=2x-8\text{ [differentiate]}\\ f\text{'}\left(x\right)=0\\ 2x-8=0\\ x=4\text{ [simplify]}\end{array}$

Now substitute in the function,

$\begin{array}{l}f\left(4\right)={\left(4\right)}^2-8\left(4\right)+9\text{ [substitute]}\\ f\left(4\right)=-7\text{ [simplify]}\end{array}$

Thus, the range of the function is $[-7,\infty )$ .

To find the zeros, use the quadratic formula,

$\begin{array}{l}f\left(x\right)=x^2-8x+9\\ a=1,b=-8,c=9\end{array}$

Use quadratic formula, substitute and simplify,

$\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(1\right)\left(9\right)}}{2\left(1\right)}\text{ [substitute]}\\ x=\frac{8\pm \sqrt{64-36}}{2}\text{ [multiply]}\\ x=\frac{8\pm 2\sqrt{7}}{2}\text{ [subtract and square root]}\\ x=4\pm \sqrt{7}\text{ [divide]}\end{array}$