Inverse Functions

Section 1.11 Inverse Functions

A function \(f: A \to B\) assigns to each element in its domain \(A\) exactly one element in its codomain \(B\text{.}\) The inverse function of \(f\text{,}\) denoted \(f^{-1}\text{,}\) is a function that "reverses" the action of \(f\text{.}\) It maps elements from the codomain back to the domain in such a way that applying \(f\) and then \(f^{-1}\) (or vice versa) returns the original input.

\begin{equation*} \text{ if } \, y = f(x), \, \text{ then } \, x=f^{-1} y \end{equation*}

Formally, \(f^{-1}\) is the inverse of \(f\) if both of these composition properties hold: \((f^{-1} \circ f)(x) = x \quad \forall_x \in A\) \((f\circ f^{-1})(y) = y \forall_y \in B \) that are in the range of \(f\) [inverse function], [Section 1.9]

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Not every function has an inverse. For a function \(f\) to have an inverse \(f^{-1}\) \(f\) must be injective (one-to-one): Different inputs produce different outputs. No two distinct \(x_1, x_2\) have \(f(x_1) = f(x_2).\) Horizontal line Test: if any horizontal line intersects the graph of \(f\) at most once.

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\(f\) must be surjective (onto) relative to its codomain: Every element in the codomain is hit by some input (i.e., the range equals the codomain).

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If both hold, \(f\) is bijective, and an inverse function exists. Inverse function graph is a mirror image of its function about \(y=x\) line. Some times, we may need to restrict the function for a perticular interval to make it one-to-one and onto so that inverse function exists.

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If a function is invertible, its inverse can be found by solving the equation \(y = f(x)\) for \(x\) in terms of \(y\text{,}\) and then swapping \(x\) and \(y\text{.}\)

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To find the inverse of a function \(y = f(x)\text{:}\)

Solve for \(x\) in terms of \(y\text{:}\) Express \(x\) as a function of \(y\text{.}\)
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Swap \(x\) and \(y\text{:}\) The resulting expression is \(y = f^{-1}(x)\text{.}\)
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Example 1.11.1.

Find inverse, domian and range of the given functions:

\begin{equation*} f(x) = 3x-5 \end{equation*}

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\begin{equation*} f(x) = \frac{2x+5}{3x-1} \end{equation*}

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\begin{equation*} f(x) = \sqrt{2x-6} \end{equation*}

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\begin{equation*} f(x) = \sqrt[3]{2x-5} \end{equation*}

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\begin{equation*} f(x) = \sqrt[3]{x-4}+1 \end{equation*}

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\begin{equation*} f(x) = e^{3x+1}-5 \end{equation*}

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Solution.

Let

\begin{equation*} y=f(x) = 3x-5 \end{equation*}

Now, interchange \(y\) to \(x\)

\begin{align*} x \amp =3y-5 \\ \therefore \, f^{-1}(x) \amp = y =\frac{x+5}{3} \end{align*}

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Let

\begin{equation*} y=\frac{2x+5}{3x-1} \end{equation*}

interchange \(y\) to \(x\)

\begin{align*} x \amp =\frac{2y+5}{3y-1} \\ or, \, 3xy-x \amp = 2y+5 \\ or, \, y(3x-2) \amp =5+x \\ \therefore \, f^{-1}(x) \amp = y =\frac{5+x}{3x-2} \end{align*}

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Let

\begin{equation*} y=\sqrt{2x-6} \end{equation*}

interchange \(y\) to \(x\)

\begin{align*} x \amp =\sqrt{2y-6} \\ or, \, x^2 \amp = 2y-6 \\ \therefore \, f^{-1}(x) \amp = y =\frac{x^2+6}{2} \end{align*}

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Domain:

\begin{equation*} 2x-6 \geq 0 \, \Rightarrow \, x\geq 3 \end{equation*}

so the domain is \([3,\infty)\)

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Range:

\begin{equation*} y =\sqrt{2x-6} \end{equation*}

Find \(x\) in terms of \(y.\)

\begin{equation*} or, \, y^2 = 2x-6 \end{equation*}

\begin{equation*} or, \, x = \frac{y^2+6}{2} \end{equation*}

This is polynomial function. Now to obey the original square root function, \(y\) should not be negative rather \(y \geq 0. \) Hence range is \([0,\infty)\)

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do yourself
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\begin{equation*} f(x) = \sqrt[3]{x-4}+1 \end{equation*}

\begin{equation*} x = \sqrt[3]{y-4}+1 \end{equation*}

\begin{equation*} x -1 = \sqrt[3]{y-4} \end{equation*}

\begin{equation*} (x -1)^3 = y-4 \end{equation*}

\begin{equation*} \therefore \, y =(x -1)^3 +4 =f^{-1}(x). \end{equation*}

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Domain: \(x\in \mathbb R\)
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Range: \(y\in \mathbb R\)
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\begin{equation*} y = e^{3x+1}-5 \end{equation*}

\begin{equation*} x = e^{3y+1}-5 \end{equation*}

\begin{equation*} x+5 = e^{3y+1} \end{equation*}

\begin{equation*} \ln{x+5} = \ln_e^{3y+1} \end{equation*}

\begin{equation*} 3y+1 = \ln{x+5} \end{equation*}

\begin{equation*} y = \frac{\ln{x+5}-1}{3} =f^{-1} (x) \end{equation*}

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Domain: \((-\infty, \infty)\)
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Range: \((-\infty, \infty)\)
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Example 1.11.2.

Find the inverse of the following functions:

\begin{equation*} y= \frac{10^x+10^{-x}}{10^x-10^{-x}} \end{equation*}

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\begin{equation*} y= \log_a{(x+\sqrt{x^2+1})} \end{equation*}

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Solution.

\begin{equation*} y= \frac{10^x+10^{-x}}{10^x-10^{-x}} \end{equation*}

swap x and y.

\begin{equation*} x= \frac{10^y+10^{-y}}{10^y-10^{-y}} = \frac{10^y+ \frac{1}{10^y}}{10^y - \frac{1}{10^y}} = \frac{10^{2y}+1}{10^{2y}-1} \end{equation*}

let \(10^{2y} =t\text{,}\) we have

\begin{equation*} x= \frac{t+1}{t-1} \end{equation*}

\begin{align*} tx-x \amp =t+1\\ t(x-1) \amp =1+x \\ or, \, t \amp =\frac{1+x}{x-1}\\ or, \, 10^{2y} \amp =\frac{1+x}{x-1}\\ or, \, \log_{10}(2y) \amp =\log_{10} \left(\frac{1+x}{x-1}\right) \\ or, \, y \amp =\frac{1}{2}\log_{10} \left(\frac{1+x}{x-1}\right) \end{align*}

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\begin{equation*} y =f(x) = \log_a{(x+\sqrt{x^2+1})} \end{equation*}

swap x and y.

\begin{align*} x \amp = \log_a{(y+\sqrt{y^2+1})} \\ a^x \amp =y+\sqrt{y^2+1} = \sqrt{y^2+1}+y \\ now, \, \frac{1}{a^x} \amp =\sqrt{y^2+1}-y\\ \text{ on solving these eqns. }, \, a^x-\frac{1}{a^x} \amp =2y\\ or, \, y \amp = \frac{1}{2}\left(a^x-a^{-x}\right) =f^{-1}(x) \end{align*}

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Example 1.11.3.

Find the inverse function of:

\begin{equation*} y=\begin{cases} x^2 & 0\leq x\leq 2 \\ 2x & x\gt 2 \end{cases} \end{equation*}

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Solution.

\begin{equation*} x=\begin{cases} y^2 & 0\leq x\leq 4 \\ 2y & x\gt 4 \end{cases} \end{equation*}

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Example 1.11.4.

Find the domain, range, and inverse of a function and sketch the graphs of function and its inverse.

\begin{equation*} f(x) = \sqrt{3-2x} \end{equation*}

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Solution.

Let

\begin{equation*} f(x) = y = \sqrt{3-2x} \end{equation*}

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For Domain:

\begin{align*} 3-2x \amp \geq 0 \\ 3 \amp \geq 2x \\ \frac{3}{2} \amp \geq x \\ x \amp \leq \frac{3}{2} \end{align*}

Hence, the domian: \((-\infty, \frac{3}{2}] \)

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For Range:

\begin{align*} y \amp = \sqrt{3-2x} \\ y^2 \amp =3-2x \\ x \amp = \frac{3-y^2}{2} \end{align*}

For \(y\) to be \(+ve\) the Range: \([0, \infty) \) As \(x\to -\infty \) \(3-2x \to \infty\) and as \(x\to \frac{3}{2} \) \(3-2x \to 0\)

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For Inverse: Interchange \(y\) and \(x.\)

\begin{equation*} x = \sqrt{3-2y} \end{equation*}

\begin{equation*} or, \, x^2 = 3-2y \end{equation*}

\begin{equation*} or, \, y = \frac{3-x^2}{2} =f^{-1}(x) \end{equation*}

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To draw graph of a function: determine its shape:

sign of x \(-ve\)
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sign of y \(+ve\)
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standard plot of \(y=\sqrt{x}\)
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Find Point (x,y): set

\begin{equation*} 3-2x=0 \end{equation*}

\begin{equation*} or, \, x=\frac{3}{2} \end{equation*}

\begin{equation*} or, \, y = \sqrt{3-2\times \frac{3}{2}} = 0 \end{equation*}

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Hence, point (x,y) = \((\frac{3}{2}, 0)\)

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Intercept:

y-intercept: set \(x=0 \)

\begin{equation*} y = \sqrt{3-2x} = \sqrt{3} \approx 1.732 \end{equation*}

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x-intercept: set \(y=0 \)

\begin{equation*} 0= \sqrt{3-2x} \end{equation*}

\begin{equation*} x=\frac{3}{2} \end{equation*}

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The final graph looks like: Remember the graph of inverse function \(f^{_1}(x) \) is a reflection of \(f(x)\) about \(y=x\) line.

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Subsection 1.11.1 Inverse Trigonometric Functions

The inverse trigonometric functions are the inverses of the basic trigonometric functions, defined on restricted domains to ensure they are one-to-one and onto. The six inverse trigonometric functions are:

(\(sin^{-1}\)): Inverse of sine function, defined on \([-\frac{\pi}{2}, \frac{\pi}{2}]\)
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(\(cos^{-1}\)): Inverse of cosine function, defined on \([0, \pi]\)
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(\(tan^{-1}\)): Inverse of tangent function, defined on \((-\frac{\pi}{2}, \frac{\pi}{2})\)
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(\(cot^{-1}\)): Inverse of cotangent function, defined on \((0, \pi)\)
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(\(sec^{-1}\)): Inverse of secant function, defined on \([0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]\)
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(\(csc^{-1}\)): Inverse of cosecant function, defined on \([-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]\)
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These functions return angles corresponding to given trigonometric ratios within their specified domains.

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