Chapter 1.2, Problem 11QR

(a)

To determine

To find:The real solution of the equation $f\left(x\right)=\sqrt{x+7}$ for $f\left(x\right)=4$ .

Answer to Problem 11QR

The real solution for the equation $f\left(x\right)=\sqrt{x+7}$ when $f\left(x\right)=4$ is $x=9$ .

Explanation of Solution

Given information:The equation is $f\left(x\right)=\sqrt{x+7}$ and $f\left(x\right)=4$ .

Calculation:

Substitute 4 for $f\left(x\right)$ in the given equation and solve.

  $\begin{array}{c}\sqrt{x+7}=4\\ {\left(\sqrt{x+7}\right)}^2=4^2\\ x+7=16\\ x=9\end{array}$

Therefore, the real solution of the equation $f\left(x\right)=\sqrt{x+7}$ for $f\left(x\right)=4$ is $x=9$ .

(b)

To determine

To find:The real solution of the equation $f\left(x\right)=\sqrt{x+7}$ for $f\left(x\right)=1$ .

Answer to Problem 11QR

The real solution of the equation $f\left(x\right)=\sqrt{x+7}$ for $f\left(x\right)=1$ is $x=-6$ .

Explanation of Solution

Given information:The equation is $f\left(x\right)=\sqrt{x+7}$ and $f\left(x\right)=1$ .

Calculation:

Substitute 1 for $f\left(x\right)$ in the given equation and solve.

  $\begin{array}{c}\sqrt{x+7}=1\\ {\left(\sqrt{x+7}\right)}^2=1^2\\ x+7=1\\ x=-6\end{array}$

Therefore, the real solution of the equation $f\left(x\right)=\sqrt{x+7}$ for $f\left(x\right)=1$ is $x=-6$ .