Question

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

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

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

Short answer

The value of

$$\begin{gathered} f\left(0\right)=\frac{-1}{4} \\ f\left(1\right)= \\ f(-1)=0 \\ f(-x)=\frac{x^2-1}{4-x} \\ -f\left(x\right)=\frac{1-x^2}{x+4} \\ f(x+1)=\frac{x(x+2)}{x+5} \\ f\left(x\right)=\frac{x^2-1}{x+4} \\ f(x+h)=\frac{{x^2+2xh+h}^2-1}{x+h+4} \end{gathered}$$

Step-by-step solution

Step 1. Given Information

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

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

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

$$\begin{gathered} f\left(0\right)=\frac{{\left(0\right)}^2-1}{0+4} \\ f\left(0\right)=\frac{-1}{4} \end{gathered}$$

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

$$\begin{gathered} f\left(1\right)=\frac{{\left(1\right)}^2-1}{1+4} \\ f\left(1\right)=\frac{1-1}{5} \\ f\left(1\right)=\frac{0}{5} \\ f\left(1\right)=0 \end{gathered}$$

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

$$\begin{gathered} f(-1)=\frac{{(-1)}^2-1}{-1+4} \\ f(-1)=\frac{1-1}{3} \\ f(-1)=\frac{0}{3} \\ f(-1)=0 \end{gathered}$$

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

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

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

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

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

$f(x+1)=\frac{{(x+1)}^2-1}{(x+1)+4}$

Using the formula ${(a+b)}^2=a^2+2ab+b^2$

$$\begin{gathered} f(x+1)=\frac{x^2+2x+1-1}{x+1+4} \\ f(x+1)=\frac{x^2+2x}{x+5} \\ f(x+1)=\frac{x(x+2)}{x+5} \end{gathered}$$

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

The value of$f\left(x\right)=\frac{x^2-1}{x+4}$

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

$f(x+h)=\frac{{(x+h)}^2-1}{(x+h)+4}$

Using the formula${(a+b)}^2=a^2+2ab+b^2$

$f(x+h)=\frac{{x^2+2xh+h}^2-1}{x+h+4}$