Question

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verifying inverses
directions: determine whether the relations are one - to - one functions.

1. graph 1
2. graph 2
3. graph 3

directions: determine if ( f(x) ) has an inverse, if yes, find ( f^{-1}(x) ). state any restrictions in the domain.

1. ( f(x)=\frac{3}{4}x + 3 )
2. ( f(x)=(x - 5)^2+9 )
3. ( f(x)=\frac{1}{x + 2} )
4. ( f(x)=sqrt3{x + 7}-1 )
5. ( f(x)=4|2x - 1| )
6. ( f(x)=\frac{2x + 5}{x - 7} )

© gina wilson (all things algebra®, llc), 2017

Explanation:

Let's solve question 4: \( f(x) = \frac{3}{4}x + 3 \)

Step 1: Check if the function is one - to - one (invertible)

A linear function of the form \( y=mx + b\) where \(m
eq0\) is a one - to - one function. For \(f(x)=\frac{3}{4}x + 3\), the slope \(m = \frac{3}{4}
eq0\), so it is one - to - one and has an inverse.

Step 2: Replace \(f(x)\) with \(y\)

We have \(y=\frac{3}{4}x + 3\)

Step 3: Swap \(x\) and \(y\)

We get \(x=\frac{3}{4}y+3\)

Step 4: Solve for \(y\)

First, subtract 3 from both sides:
\(x - 3=\frac{3}{4}y\)
Then, multiply both sides by \(\frac{4}{3}\) to isolate \(y\):
\(y=\frac{4}{3}(x - 3)=\frac{4}{3}x-4\)

Step 5: Replace \(y\) with \(f^{-1}(x)\)

So, \(f^{-1}(x)=\frac{4}{3}x - 4\)

The domain of the original function \(f(x)=\frac{3}{4}x + 3\) is all real numbers (\((-\infty,\infty)\)) since it is a linear function. There are no restrictions on the domain for the inverse function \(f^{-1}(x)\) other than the domain being all real numbers as well (because the range of \(f(x)\) is all real numbers).

Answer:

The function \(f(x)=\frac{3}{4}x + 3\) has an inverse. The inverse function is \(f^{-1}(x)=\frac{4}{3}x - 4\) and the domain of \(f(x)\) (and the range of \(f^{-1}(x)\)) is \((-\infty,\infty)\), and the domain of \(f^{-1}(x)\) (and the range of \(f(x)\)) is \((-\infty,\infty)\)