Question

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

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

localid="1645960232226" $f\left(x\right)=\frac{x}{x^2+1}$

Short answer

The value of

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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