Chapter 7.7, Problem 5ST

To determine

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

Answer to Problem 5ST

The domain, range and zeros of the function is $R,[-6,\infty )\text{ and }x=3\pm \sqrt{6}$ .

Explanation of Solution

Given information:

The function is $g\left(x\right)=x^2-6x+3$ .

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

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

To find the range differentiate the function, and put equal to zero

$\begin{array}{l}g\left(x\right)=x^2-6x+3\\ g\text{'}\left(x\right)=2x-6\text{ [differentiate]}\\ 2x-6=0\text{ [equal to zero]}\\ x=3\text{ [simplify]}\end{array}$

Now, substitute in the function,

$\begin{array}{l}g\left(3\right)={\left(3\right)}^2-6\left(3\right)+3\text{ [substitute]}\\ g\left(3\right)=-6\text{ [simplify]}\end{array}$

Thus, the range of the function is $[-6,\infty )$ .

To find the zeros, put the function equal to zero,

$\begin{array}{l}{x}^2-6x+3=0\\ x=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(3\right)}}{2\left(1\right)}\text{ [use the quadratic formula]}\\ x=\frac{6\pm \sqrt{36-12}}{2}\text{ [square and multiply]}\\ x=3\pm \sqrt{6}\text{ [simplify]}\end{array}$

Thus, The domain, range and zeros of the function is $R,[-6,\infty )\text{ and }x=3\pm \sqrt{6}$ .