Question

examples 2 and 3
find the inverse of each function. then graph th
necessary, restrict the domain of the inverse so

1. ( f(x) = x + 2 )
2. ( f(x) = -2x + 1 )
3. ( f(x) = -\frac{5}{3}x - 8 )

Explanation:

Problem 5: \( f(x) = x + 2 \)

Step 1: Replace \( f(x) \) with \( y \)

We start by writing the function as \( y = x + 2 \).

Step 2: Swap \( x \) and \( y \)

To find the inverse, we interchange \( x \) and \( y \), so we get \( x = y + 2 \).

Step 3: Solve for \( y \)

Subtract 2 from both sides of the equation: \( y = x - 2 \).

Step 4: Replace \( y \) with \( f^{-1}(x) \)

The inverse function is \( f^{-1}(x) = x - 2 \). Since the original function is a linear function (and thus one - to - one over its entire domain \( (-\infty, \infty) \)), the inverse function has the domain \( (-\infty, \infty) \) and we don't need to restrict the domain.

Step 1: Replace \( f(x) \) with \( y \)

We have \( y=-2x + 1 \).

Step 2: Swap \( x \) and \( y \)

Interchanging \( x \) and \( y \) gives us \( x=-2y + 1 \).

Step 3: Solve for \( y \)

First, subtract 1 from both sides: \( x - 1=-2y \). Then divide both sides by - 2: \( y=\frac{-x + 1}{2}=-\frac{1}{2}x+\frac{1}{2}\).

Step 4: Replace \( y \) with \( f^{-1}(x) \)

The inverse function is \( f^{-1}(x)=-\frac{1}{2}x+\frac{1}{2}\). The original function is a linear function (one - to - one over \( (-\infty, \infty) \)), so the inverse function has the domain \( (-\infty, \infty) \) and no domain restriction is needed.

Step 1: Replace \( f(x) \) with \( y \)

We get \( y =-\frac{5}{3}x-8 \).

Step 2: Swap \( x \) and \( y \)

Interchanging \( x \) and \( y \) gives \( x=-\frac{5}{3}y-8 \).

Step 3: Solve for \( y \)

First, add 8 to both sides: \( x + 8=-\frac{5}{3}y \). Then multiply both sides by \( -\frac{3}{5} \): \( y=-\frac{3}{5}(x + 8)=-\frac{3}{5}x-\frac{24}{5}\).

Step 4: Replace \( y \) with \( f^{-1}(x) \)

The inverse function is \( f^{-1}(x)=-\frac{3}{5}x-\frac{24}{5}\). The original function is a linear function (one - to - one over \( (-\infty, \infty) \)), so the inverse function has the domain \( (-\infty, \infty) \) and no domain restriction is needed.

Answer:

The inverse of \( f(x)=x + 2\) is \( f^{-1}(x)=x - 2\)

Problem 7: \( f(x)=-2x + 1 \)