Chapter 1, Problem 46RE

a.

To determine

To find : the value of independent variable and simplify.

Answer to Problem 46RE

the value of $f\left(1\right)$ is $-1$.

Explanation of Solution

Given information :

The function is $f\left(x\right)=\frac{3}{2x-5}$.

  $f\left(1\right)$

Calculation: since here $x=1$ ,so first substitute then multiply and subtract then divide,

  $\begin{array}{l}f\left(x\right)=\frac{3}{2x-5}\\ f\left(1\right)=\frac{3}{2\left(1\right)-5}\text{ [substitute }x=1\text{]}\\ f\left(1\right)\text{=}\frac{3}{2-5}\text{ [multiply]}\\ f\left(1\right)=-1\text{ [subtract and divide]}\end{array}$

The value is $-1$.

b.

To determine

To find : the value of independent variable and simplify.

Answer to Problem 46RE

the value of $f\left(-2\right)$ is $-\frac{1}{3}$.

Explanation of Solution

Given information :

The function is $f\left(x\right)=\frac{3}{2x-5}$.

  $f\left(-2\right)$

Calculation: since here $x=-2$ ,so first substitute then, multiply and add then divide,

  $\begin{array}{l}f\left(x\right)=\frac{3}{2x-5}\\ f\left(-2\right)=\frac{3}{2\left(-2\right)-5}\text{ [substitute }x=-2\text{]}\\ f\left(-2\right)\text{=}\frac{3}{-4-5}\text{ [multiply]}\\ f\left(-2\right)=-\frac{1}{3}\text{ [add and divide]}\end{array}$

The value is $-\frac{1}{3}$.

c.

To determine

To find : the value of independent variable and simplify.

Answer to Problem 46RE

the value of $f\left(t\right)$ is $\frac{3}{2t-5}$.

Explanation of Solution

Given information :

The function is $f\left(x\right)=\frac{3}{2x-5}$.

  $f\left(t\right)$

Calculation: since here $x=t$ ,so first substitute then multiply,

  $\begin{array}{l}f\left(x\right)=\frac{3}{2x-5}\\ f\left(t\right)=\frac{3}{2\left(t\right)-5}\text{ [substitute }x=t\text{]}\\ f\left(t\right)\text{=}\frac{3}{2t-5}\text{ [multiply]}\end{array}$

The value is $\frac{3}{2t-5}$.

d.

To determine

To find : the value of independent variable and simplify.

Answer to Problem 46RE

the value of $f\left(10\right)$ is $-\frac{1}{5}$.

Explanation of Solution

Given information :

The function is $f\left(x\right)=\frac{3}{2x-5}$.

  $f\left(10\right)$

Calculation: since here $x=10$ ,so first substitute then square and add,

  $\begin{array}{l}f\left(x\right)=\frac{3}{2x-5}\\ f\left(10\right)=\frac{3}{2\left(10\right)-5}\text{ [substitute }x=10\text{]}\\ f\left(10\right)\text{=}\frac{3}{20-5}\text{ [multiply]}\\ f\left(10\right)=-\frac{1}{5}\text{ [subtract and divide]}\end{array}$

The value is $-\frac{1}{5}$.