The inverse function \(\displaystyle f^{-1}(x)\) of a function \(\displaystyle f(x)\) can be found as follows:

Replace \(\displaystyle f(x)\) with \(\displaystyle y\):

\(\displaystyle f(x) = \sqrt{2x- 5}\)

\(\displaystyle y= \sqrt{2x- 5}\)

Switch the positions of \(\displaystyle y\) and \(\displaystyle x\):

\(\displaystyle x= \sqrt{2y- 5}\),

or

\(\displaystyle \sqrt{2y- 5} = x\)

Solve for \(\displaystyle y\). This can be done as follows:

Square both sides:

\(\displaystyle \left (\sqrt{2y- 5} \right ) ^{2} = x^{2}\)

\(\displaystyle 2y- 5 = x^{2}\)

Add 5 to both sides:

\(\displaystyle 2y- 5 + 5 = x^{2} + 5\)

\(\displaystyle 2y = x^{2} + 5\)

Multiply both sides by \(\displaystyle \frac{1}{2}\), distributing on the right:

\(\displaystyle \frac{1}{2} \cdot 2y =\frac{1}{2} \cdot \left ( x^{2} + 5 \right )\)

\(\displaystyle y =\frac{1}{2} \cdot x^{2} + \frac{1}{2} \cdot 5\)

\(\displaystyle y =\frac{1}{2} x^{2} + \frac{5}{2}\)

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

\(\displaystyle f^{-1}(x) =\frac{1}{2} x^{2} + \frac{5}{2}\),

the correct response.

Given the graph of \(\displaystyle f(x)\), the graph of its inverse, \(\displaystyle f^{-1}(x)\) is the reflection of the former about the line \(\displaystyle y = x\). This line is in dark green below; critical points are reflected as shown:

The figure in blue is the graph of \(\displaystyle f^{-1}(x)\)

A relation is a function if and only if it passes the Vertical Line Test (VLT) - that is, no vertical line exists that passes through its graph more than once. From the diagram below, we see that no such line exists:

The relation passes the VLT, so it is a function.

A function has an inverse if and only if it passes the Horizontal Line Test (HLT) - that is, no horizontal line exists that passes through its graph more than once. From the diagram below, we see that no such line exists:

The function passes the HLT, so it has an inverse.

The inverse function \(\displaystyle f^{-1}(x)\) of a function \(\displaystyle f(x)\) can be found as follows:

Replace \(\displaystyle f(x)\) with \(\displaystyle y\):

\(\displaystyle f(x) = \ln (3x+ 7)\)

\(\displaystyle y= \ln (3x+ 7)\)

Switch the positions of \(\displaystyle y\) and \(\displaystyle x\):

\(\displaystyle x= \ln (3y+ 7)\)

or

\(\displaystyle \ln (3y+ 7) = x\)

Solve for \(\displaystyle y\) - that is, isolate it on one side - as follows:

Raise \(\displaystyle e\) to the power of both sides:

\(\displaystyle e^{ \ln (3y+ 7)} = e^{x}\)

A property of logarithms states that \(\displaystyle e ^{\log N} = N\), so

\(\displaystyle 3y+ 7 = e^{x}\)

Subtract 7 from both sides:

\(\displaystyle 3y+ 7 -7 = e^{x} - 7\)

\(\displaystyle 3y = e^{x} - 7\)

Multiply both sides by \(\displaystyle \frac{1}{3}\), distributing on the right:

\(\displaystyle \frac{1}{3} \cdot 3y =\frac{1}{3} \cdot( e^{x} - 7)\)

\(\displaystyle y =\frac{1}{3} \cdot e^{x} - \frac{1}{3} \cdot7\)

\(\displaystyle y =\frac{1}{3} e^{x} - \frac{7}{3}\)

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

\(\displaystyle f^{-1}(x) =\frac{1}{3} e^{x} - \frac{7}{3}\)

A relation is a function if and only if it passes the Vertical Line Test (VLT)—that is, no vertical line exists that passes through its graph more than once. From the diagram below, we see that no such line exists:

A function has an inverse, if and only if, it passes the Horizontal Line Test (HLT) - that is, no horizontal line exists that passes through its graph more than once. From the diagram below, we see at least one such line exists.

One definition of the inverse function \(\displaystyle f^{-1}(x)\) of a function \(\displaystyle f(x)\) is the set of all ordered pairs \(\displaystyle (b, a)\) such that the ordered pair \(\displaystyle (a,b)\) is in the set of ordered pairs in \(\displaystyle f(x)\). As such, if the ordered pairs of \(\displaystyle f(x)\) are given, as is the case here, the set of ordered pairs in \(\displaystyle f^{-1}(x)\) can be found by switching the positions of the coordinates in all of the pairs. Doing this, we obtain:

or, ordering the \(\displaystyle x\)-coordinates,

A relation is a function, if and only if, it passes the Vertical Line Test (VLT)—that is, no vertical line exists that passes through its graph more than once. From the diagram below, we see that no such line exists:

The relation passes the VLT, so it is a function.

A function has an inverse if and only if it passes the Horizontal Line Test (HLT) - that is, no horizontal line exists that passes through its graph more than once. From the diagram below, we see that no such line exists:

The function passes the HLT, so it has an inverse.

A relation is a function, if and only if, it passes the Vertical Line Test (VLT)—that is, no vertical line exists that passes through its graph more than once. From the diagram below, we see that no such line exists;

The relation passes the VLT, so it is a function.

A function has an inverse if and only if it passes the Horizontal Line Test (HLT) - that is, no horizontal line exists that passes through its graph more than once. From the diagram below, we see at least one such line exists.

The function fails the HLT, so it does not have an inverse.

To find the inverse of a linear function, switch the variables and solve for y.

\(\displaystyle y=\frac{x-3}{2}\)

Switch the variables:

\(\displaystyle x=\frac{y-3}{2}\)

Multiply both sides by 2:

\(\displaystyle 2x=y-3\)

Add 3 to both sides to isolate y:

Interchange the x and y-variables.

\(\displaystyle x=2(5-y)\)

Solve for y.  Divide by two on both sides.

\(\displaystyle \frac{x}{2}=\frac{2(5-y)}{2}\)

\(\displaystyle \frac{x}{2}=5-y\)

Add \(\displaystyle y\) on both sides.

\(\displaystyle \frac{x}{2}+y=5-y+y\)

\(\displaystyle \frac{x}{2}+y=5\)

Subtract \(\displaystyle \frac{x}{2}\) on both sides.

\(\displaystyle \frac{x}{2}+y-\frac{x}{2}=5-\frac{x}{2}\)

Simplify both sides.

The answer is: