Question

A function $f:[-3,7)\to \text{R}$ is defined as follows:
$f\left(x\right)=$$(\begin{array}{c}4x^2-1;-3\le x<2\\ 3x-2;2\le x\le 4\\ 2x-3;4<x<7\end{array}$

Find: $\frac{f\left(3\right)+f(-1)}{2f\left(6\right)-f\left(1\right)}$

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{4}$
• D. $\frac{4}{5}$

Answer 1

The correct option is B $\frac{2}{3}$

$f\left(3\right)=3x–2=3\left(3\right)–2=9–2=7$

$f(-1)=4x^2–1=4(-1{)}^2–1=$$4-1=3$

$f\left(6\right)=2x–3=2\left(6\right)–3=12–3=9$

$f\left(1\right)=4x^2–1=4(1{)}^2–1=$$4-1=3$

$\frac{f\left(3\right)+f(-1)}{2f\left(6\right)-f\left(1\right)}=\frac{7+3}{2\left(9\right)-3}=\frac{10}{15}=\frac{2}{3}$