Question

Find the following for each function:

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

Part (a):$f\left(2\right)$

Part (b):$f\left(-2\right)$

Part (c):$f\left(-x\right)$

Part (d):$-f\left(x\right)$

Part (e):$f\left(x-2\right)$

Part (f):$f\left(2x\right)$

Short answer

Part (a):$f\left(2\right)=2$

Part (b): $f\left(-2\right)=-2$

Part (c):$f\left(-x\right)=\frac{-3x}{x^2-1}$

Part (d):$-f\left(x\right)=\frac{-3x}{-x^2+1}$

Part (e):$f\left(x-2\right)=\frac{3\left(x-2\right)}{x^2-4x+3}$

Part (f):$f\left(2x\right)=\frac{6x}{4x^2-1}$

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given function,

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

Substitute $x=2$in the given equation,

localid="1645858263721" $$\begin{gathered} f\left(2\right)=\frac{3\left(2\right)}{{\left(2\right)}^2-1} \\ f\left(2\right)=\frac{6}{4-1} \\ f\left(2\right)=\frac{6}{3} \\ f\left(2\right)=2 \end{gathered}$$

Part (b) Step 1. Given information.

Consider the given function,

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

Substitute $x=-2$in the given equation,

$$\begin{gathered} f\left(-2\right)=\frac{3\left(-2\right)}{{\left(-2\right)}^2-1} \\ f\left(-2\right)=\frac{-6}{4-1} \\ f\left(-2\right)=\frac{-6}{3} \\ f\left(-2\right)=-2 \end{gathered}$$

Part (c) Step 1. Given information.

Consider the given function,

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

Substitute $x=-x$in the given equation,

$$\begin{gathered} f\left(-x\right)=\frac{3\left(-x\right)}{{\left(-x\right)}^2-1} \\ f\left(-x\right)=\frac{-3x}{x^2-1} \end{gathered}$$

Part (d) Step 1. Given information.

Consider the given function,

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

From the question,

$$\begin{gathered} -f\left(x\right)=-\left(\frac{3x}{x^2-1}\right) \\ -f\left(x\right)=\frac{-3x}{-x^2+1} \end{gathered}$$

Part (e) Step 1. Given information.

Consider the given function,

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

Substitute $x-2$in the given equation,

$$\begin{gathered} f\left(x-2\right)=\frac{3\left(x-2\right)}{{\left(x-2\right)}^2-1} \\ f\left(x-2\right)=\frac{3\left(x-2\right)}{x^2+4-4x-1} \\ f\left(x-2\right)=\frac{3\left(x-2\right)}{x^2-4x+3} \end{gathered}$$

Part (f) Step 1. Given information.

Consider the given function,

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

Substitute $2x$in the given equation,

$$\begin{gathered} f\left(2x\right)=\frac{3\left(2x\right)}{{\left(2x\right)}^2-1} \\ f\left(2x\right)=\frac{6x}{4x^2-1} \end{gathered}$$