Question

In Problems 39– 46, find the following for each function:

$\left(a\right)f\left(0\right)\left(b\right)f\left(1\right)\left(c\right)f(-1)\left(d\right)f(-x)\left(e\right)-f\left(x\right)\left(f\right)f(x+1)\left(g\right)f\left(x\right)\left(h\right)f(x+h)$

$f\left(x\right)=\frac{2x+1}{3x-5}$'

Short answer

The value of

$$\begin{gathered} f\left(0\right)=\frac{1}{-5} \\ f\left(1\right)=\frac{3}{-2} \\ f(-1)=\frac{1}{8} \\ f(-x)=\frac{-2x+1}{-3x-5} \\ -f\left(x\right)=-\frac{2x+1}{3x-5} \\ f(x+1)=\frac{2x+3}{3x-2} \\ f\left(x\right)=\frac{2x+1}{3x-5} \\ f(x+h)=\frac{2x+2h+1}{3x+3h-5} \end{gathered}$$

Step-by-step solution

Step 1. Given Information

We have to find the value for each function:

$\left(a\right)f\left(0\right)\left(b\right)f\left(1\right)\left(c\right)f(-1)\left(d\right)f(-x)\left(e\right)-f\left(x\right)\left(f\right)f(x+1)\left(g\right)f\left(x\right)\left(h\right)f(x+h)$

The given equation is$f\left(x\right)=\frac{2x+1}{3x-5}$

Part (a) Step 1. We find the value of  f(0)Putting x=0 in the given question.

$$\begin{gathered} f\left(0\right)=\frac{2\times 0+1}{3\times 0-5} \\ f\left(0\right)=\frac{1}{-5} \end{gathered}$$

Part (b) Step 1. We find the value of  f(1)Putting x=1 in the given question.

$$\begin{gathered} f\left(1\right)=\frac{2\times 1+1}{3\times 1-5} \\ f\left(1\right)=\frac{2+1}{3-5} \\ f\left(1\right)=\frac{3}{-2} \end{gathered}$$

Part (c) Step 1. We find the value of f(-1)Putting x=-1 in the given question.

$$\begin{gathered} f(-1)=\frac{2\times (-1)+1}{3\times (-1)-5} \\ f(-1)=\frac{-2+1}{-3-5} \\ f(-1)=\frac{-1}{-8} \\ f(-1)=\frac{1}{8} \end{gathered}$$

Part (d) Step 1. We find the value of f(-x)Putting x=-x in the given question.

$$\begin{gathered} f(-x)=\frac{2(-x)+1}{3(-x)-5} \\ f(-x)=\frac{-2x+1}{-3x-5} \end{gathered}$$

Part (e) Step 1. We find the value of -f(x)To find the value we multiply the given equation by -sign

$-f\left(x\right)=-\frac{2x+1}{3x-5}$

Part (f) Step 1. We find the value of f(x+1)Putting x=x+1 in the given question.

$$\begin{gathered} f(x+1)=\frac{2(x+1)+1}{3(x+1)-5} \\ f(x+1)=\frac{2x+2+1}{3x+3-5} \\ f(x+1)=\frac{2x+3}{3x-2} \end{gathered}$$

Part (g) Step 1. We find the value of f(x)

The value of$f\left(x\right)=\frac{2x+1}{3x-5}$

Part (h) Step 1. We find the value of f(x+h)Putting x=x+h in the given question.

$$\begin{gathered} f(x+h)=\frac{2(x+h)+1}{3(x+h)-5} \\ f(x+h)=\frac{2x+2h+1}{3x+3h-5} \end{gathered}$$