Based on the figure above, inverse function can be conclude as:

\(f:x\rightarrow y\Leftrightarrow f^{-1}:y\rightarrow x\) or \(y=f(x) \Leftrightarrow x=f^{-1}(y)\).

To determine whether the graph of a function has an inverse function, carry out the horizontal line test. If the horizontal line cuts the graph of the function at only one point, then this type of function is a one-to-one function and it has an inverse function. Conversely, if the horizontal line cuts the graph at two or more points, then this type of function is not a one-to- one and the function has no inverse function.

A function \(f\) that maps set \(X\) to set \(Y\) has an inverse function, \(f^{-1}\) if \(f\) is a one-to-one function.

\(fg(x)=x\), \(x\) in the domain of \(g\) and \(gf(x)=x\), \(x\) in the domain of \(f\).

If two functions \(f\) and \(g\) are inverse functions of each other, then

(a) the domain of \(f=\) range of \(g\) and domain of \(g=\) range of \(f\),
(b) the graph of \(g\) is the reflection of the graph of \(f\) at the line \(y=x\).

For any real number, \(a\) and \(b\), if the point \((a,b)\) lies on the graph \(f\), then the point \((b,a)\) lies on the graph \(g\), that is graph \(f^{-1}\). The point \((b,a)\) lies on the graph \(g\) is the point of reflection of \((a,b)\) which lies on the graph \(f\) in the line \(y=x\).

Verify that the function \(f(x)=3-2x\) has an inverse function, \(g(x)=\dfrac{3-x}{2}\).

First, determine \(fg(x)\),

\(\begin{aligned} fg(x)&=f[g(x)] \\\\ &=f\left( \dfrac{3-x}{2} \right) \\\\ &=3-2\left( \dfrac{3-x}{2} \right) \\\\ &=3-(3-x) \\\\ &=x .\end{aligned}\)

Then, determine \(gf(x)\).

\(\begin{aligned} gf(x)&=g[f(x)] \\\\ &=g(3-2x) \\\\ &=\dfrac{3-(3-2x)}{2} \\\\ &=\dfrac{2x}{2} \\\\ &=x. \end{aligned}\)

Since \(fg(x)=gf(x)=x\),

thus,

\(g(x)=\dfrac{3-x}{2}\) is the inverse function of \(f(x)=3-2x\).

If \(f(x)=5x+2\), find

(a) \(f^{-1}(x)\),
(b) \(f^{-1}(7)\).

(a)

Let \(f(x)=y\),

\(\begin{aligned} y&=5x+2\\\\ 5x&=y-2 \\\\ x&=\dfrac{y-2}{5} \end{aligned}\)

thus,

\(f^{-1}(x)=\dfrac{x-2}{5}.\)

(b)

Substitute the value \(7\) into \(f^{-1}(x)=\dfrac{x-2}{5}\),

\(\begin{aligned} f^{-1}(7)&=\dfrac{7-2}{5}\\\\ &=\dfrac{5}{5}\\\\ &=1. \end{aligned}\)

\(\therefore f^{-1}(7)=1.\)

Based on the figure above, inverse function can be conclude as:

\(f:x\rightarrow y\Leftrightarrow f^{-1}:y\rightarrow x\) or \(y=f(x) \Leftrightarrow x=f^{-1}(y)\).

To determine whether the graph of a function has an inverse function, carry out the horizontal line test. If the horizontal line cuts the graph of the function at only one point, then this type of function is a one-to-one function and it has an inverse function. Conversely, if the horizontal line cuts the graph at two or more points, then this type of function is not a one-to- one and the function has no inverse function.

A function \(f\) that maps set \(X\) to set \(Y\) has an inverse function, \(f^{-1}\) if \(f\) is a one-to-one function.

\(fg(x)=x\), \(x\) in the domain of \(g\) and \(gf(x)=x\), \(x\) in the domain of \(f\).

If two functions \(f\) and \(g\) are inverse functions of each other, then

(a) the domain of \(f=\) range of \(g\) and domain of \(g=\) range of \(f\),
(b) the graph of \(g\) is the reflection of the graph of \(f\) at the line \(y=x\).

For any real number, \(a\) and \(b\), if the point \((a,b)\) lies on the graph \(f\), then the point \((b,a)\) lies on the graph \(g\), that is graph \(f^{-1}\). The point \((b,a)\) lies on the graph \(g\) is the point of reflection of \((a,b)\) which lies on the graph \(f\) in the line \(y=x\).

Verify that the function \(f(x)=3-2x\) has an inverse function, \(g(x)=\dfrac{3-x}{2}\).

First, determine \(fg(x)\),

\(\begin{aligned} fg(x)&=f[g(x)] \\\\ &=f\left( \dfrac{3-x}{2} \right) \\\\ &=3-2\left( \dfrac{3-x}{2} \right) \\\\ &=3-(3-x) \\\\ &=x .\end{aligned}\)

Then, determine \(gf(x)\).

\(\begin{aligned} gf(x)&=g[f(x)] \\\\ &=g(3-2x) \\\\ &=\dfrac{3-(3-2x)}{2} \\\\ &=\dfrac{2x}{2} \\\\ &=x. \end{aligned}\)

Since \(fg(x)=gf(x)=x\),

thus,

\(g(x)=\dfrac{3-x}{2}\) is the inverse function of \(f(x)=3-2x\).

If \(f(x)=5x+2\), find

(a) \(f^{-1}(x)\),
(b) \(f^{-1}(7)\).

(a)

Let \(f(x)=y\),

\(\begin{aligned} y&=5x+2\\\\ 5x&=y-2 \\\\ x&=\dfrac{y-2}{5} \end{aligned}\)

thus,

\(f^{-1}(x)=\dfrac{x-2}{5}.\)

(b)

Substitute the value \(7\) into \(f^{-1}(x)=\dfrac{x-2}{5}\),

\(\begin{aligned} f^{-1}(7)&=\dfrac{7-2}{5}\\\\ &=\dfrac{5}{5}\\\\ &=1. \end{aligned}\)

\(\therefore f^{-1}(7)=1.\)