Inverse functions

Exercises · Functions · with worked solutions · PDF

Exercise 1

Prove that the function

\[ f(x)=\log{(2+3x)} \]

is invertible on its domain and determine the analytic expression of the inverse \(f^{-1}\). Specify the domain of \(f^{-1}\).

Solution

The domain of the given function is

\[ \left(-\frac{2}{3},+\infty\right), \]

the interval where the argument of the logarithm is strictly positive. To prove that \(f\) is invertible, it suffices to verify that it is injective, i.e.,

\[ \forall x_1,x_2 \in D \qquad f(x_1) = f(x_2) \Longrightarrow x_1 = x_2 \]

Imposing \(f(x_1) = f(x_2)\) is equivalent to imposing that

\[\begin{align*} \log{(2+3x_1)} &=\log{(2+3x_2)} \\ 2+3x_1 &= 2+3x_2 \\ 3x_1 &= 3x_2 \Longleftrightarrow x_1=x_2 \end{align*}\]

The function \(f\) is injective and, consequently, invertible. We look for the analytic expression of the inverse function \(f^{-1}\) by solving, with respect to the variable \(x\), the equation

\[ \log{(2+3x)}=y. \]

which has the unique solution

\[ x=\frac{1}{3}(e^{y}-2) \]

Solution

More formally, the inverse function of \(f\) is

\[ f^{-1}(y)=\frac{1}{3}(e^{y}-2) \]

whose domain is the image of the original function \(f\), that is, \(\mathbb{R}\). Hence \(f:\left(-\frac{2}{3},+\infty\right)\to(-\infty, +\infty)\), while \(f^{-1}:(-\infty, +\infty) \to\left(-\frac{2}{3},+\infty\right)\).

The graphs \(y=\log{(2+3x)}\) and \(y=\frac{1}{3}(e^{x}-2)\) are symmetric with respect to the bisector \(y=x\):

Exercise 2

Prove that the function

\[ f(x)=\sqrt{1-2x} \]

is invertible on its domain and determine the analytic expression of the inverse \(f^{-1}\). Specify the domain of \(f^{-1}\).

Solution

The domain of the given function is

\[ \left(-\infty,\frac{1}{2}\right], \]

the interval where the radicand is non-negative.

Solution

To prove that \(f\) is invertible, it suffices to verify that it is injective, i.e.,

\[ \forall x_1,x_2 \in D \qquad f(x_1) = f(x_2) \Longrightarrow x_1 = x_2 \]

Imposing \(f(x_1) = f(x_2)\) is equivalent to imposing that

\[\begin{align*} \sqrt{1-2x_1} &=\sqrt{1-2x_2} \\ 1-2x_1 &=1-2x_2 \\ -2x_1 &= -2x_2 \Longleftrightarrow x_1=x_2 \end{align*}\]

The function \(f\) is injective and, consequently, invertible. We look for the analytic expression of the inverse function \(f^{-1}\) by solving, with respect to the variable \(x\), the equation

\[ \sqrt{1-2x}=y. \]

which, for \(y \geq 0\) (a square root is never negative in \(\mathbb{R}\)), has the unique solution

\[ x=-\frac{1}{2}y^2+\frac{1}{2} \]

More formally, the inverse function of \(f\) is

\[ f^{-1}(y)=-\frac{1}{2}y^2+\frac{1}{2} \]

whose domain is the image of the original function \(f\), that is, \([0, +\infty)\). Hence \(f:\left(-\infty,\frac{1}{2}\right]\to[0, +\infty)\), \(f^{-1}:[0, +\infty) \to\left(-\infty,\frac{1}{2}\right]\).

The graphs \(y=\sqrt{1-2x}\) and \(y=-\frac{1}{2}x^2+\frac{1}{2}\) are symmetric with respect to the bisector \(y=x\):

Exercise 3

Prove that the function

\[ f(x)=e^{2x-3} \]

is invertible on its domain and determine the analytic expression of the inverse \(f^{-1}\). Specify the domain of \(f^{-1}\).

Solution

The domain of the given function is

\[ \left(-\infty,+\infty\right), \]

since the exponential function is defined for every \(x\in\mathbb{R}\). To prove that \(f\) is invertible, it suffices to verify that it is injective, i.e.,

\[ \forall x_1,x_2 \in D \qquad f(x_1) = f(x_2) \Longrightarrow x_1 = x_2 \]

Imposing \(f(x_1) = f(x_2)\) is equivalent to imposing that

\[\begin{align*} e^{2x_1-3} &=e^{2x_2-3} \\ 2x_1-3 &=2x_2-3 \\ 2x_1 &= 2x_2 \Longleftrightarrow x_1=x_2 \end{align*}\]

The function \(f\) is injective and, consequently, invertible. We look for the analytic expression of the inverse function \(f^{-1}\) by solving, with respect to the variable \(x\), the equation

\[ e^{2x-3}=y. \]

It makes sense to solve this equation only for \(y > 0\), since it would have no solutions when \(y\leq0\) (the exponential never vanishes and is never negative). The equation has the unique solution

\[ x=\frac{3}{2}+\frac{1}{2}\log{y} \]

More formally, the inverse function of \(f\) is

\[ f^{-1}(y)=\frac{3}{2}+\frac{1}{2}\log{y} \]

whose domain is the image of the original function \(f\), that is, \((0, +\infty)\). Hence \(f:(-\infty,+\infty)\to(0, +\infty)\), \(f^{-1}:(0, +\infty) \to (-\infty,+\infty)\).

Exercise 4

Prove that the function

\[ f(x)=\arctan(2x-1) \]

is invertible on its domain and determine the analytic expression of the inverse \(f^{-1}\). Specify the domain of \(f^{-1}\).

Solution

The arctangent function, which we recall is the inverse function of the restriction of \(y=\tan(x)\) to the interval \((-\frac{\pi}{2},\frac{\pi}{2})\), has the following properties:

• its domain is the set \(\mathbb{R}\);

• its image is the interval \((-\frac{\pi}{2},\frac{\pi}{2})\);

• it is a strictly increasing monotonic function.

Since it is strictly increasing, the function \(f\) is invertible (note that this is only a sufficient condition). We look for the analytic expression of the inverse function \(f^{-1}\) by solving, with respect to the variable \(x\), the equation

\[ \arctan(2x-1)=y, \]

where \(y \in (-\frac{\pi}{2},\frac{\pi}{2})\). The equation becomes

\[\begin{align*} 2x-1 &=\tan(y) \\ x &= \frac{1}{2}+\frac{\tan(y)}{2} \end{align*}\]

More formally, the inverse function of \(f\) is

\[ f^{-1}(y)=\frac{1}{2}+\frac{\tan(y)}{2} \]

whose domain is the image of the original function \(f\), that is, \((-\frac{\pi}{2},\frac{\pi}{2})\). Hence \(f:\mathbb{R}\to(-\frac{\pi}{2},\frac{\pi}{2})\), while \(f^{-1}:(-\frac{\pi}{2},\frac{\pi}{2})\to\mathbb{R}\)

Exercise 5

Prove that the function

\[ f(x)=\frac{x-1}{x+2} \]

is invertible on its domain and determine the analytic expression of the inverse \(f^{-1}\). Specify the domain of \(f^{-1}\).

Solution

The domain of the given function is

\[ (-\infty, -2)\cup(-2,+\infty), \]

since it is defined \(\forall x \in \mathbb{R}, \, x \neq -2\). To prove that \(f\) is invertible, it suffices to verify that it is injective, i.e.,

\[ \forall x_1,x_2 \in D \qquad f(x_1) = f(x_2) \Longrightarrow x_1 = x_2 \]

To simplify the subsequent computations, it is convenient to rewrite the function \(f\) as:

\[ f(x)=\frac{x-1}{x+2}=\frac{x+2-3}{x+2}=1-\frac{3}{x+2} \]

Imposing \(f(x_1) = f(x_2)\) is equivalent to imposing that

\[\begin{align*} 1-\frac{3}{x_1+2} &=1-\frac{3}{x_2+2} \\ \frac{3}{x_1+2} &=\frac{3}{x_2+2} \\ x_1+2 &= x_2+2 \Longleftrightarrow x_1=x_2 \end{align*}\]

The function \(f\) is injective and, consequently, invertible. We look for the analytic expression of the inverse function \(f^{-1}\) by solving, with respect to the variable \(x\), the equation

\[ \frac{x-1}{x+2}=y. \]

Rewriting the function again as before

\[ 1-\frac{3}{x+2}=y \]

we obtain

\[ x=\frac{3}{1-y}-2=\frac{2y+1}{1-y} \]

More formally, the inverse function of \(f\) is

\[ f^{-1}(y)=\frac{2y+1}{1-y} \]

whose domain is the image of the original function \(f\), that is, \((-\infty, 1)\cup(1,+\infty)\). Hence \(f:(-\infty, -2)\cup(-2,+\infty)\to(-\infty, 1)\cup(1,+\infty)\), \(f^{-1}:(-\infty, 1)\cup(1,+\infty) \to (-\infty, -2)\cup(-2,+\infty)\).

Exercise 6

Prove that the function

\[ f(x)=e^{1+x^2} \]

is invertible on a restriction of its domain and determine the analytic expression of the inverse \(f^{-1}\). Specify the domain of \(f^{-1}\).

Solution

The domain of the given function is

\[ \left(-\infty,+\infty\right), \]

since the exponential function is defined for every \(x\in\mathbb{R}\). It is clear that \(f\) is not invertible on all of \(\mathbb{R}\), since it is an even function, i.e., symmetric with respect to the \(y\)-axis:

\[ f(-x)=e^{1+(-x)^2}=e^{1+x^2}=f(x) \]

This implies that any horizontal line intersects the graph of the function in two points. The function is not injective, hence not invertible on \(\mathbb{R}\). However, we can restrict the domain by considering only the positive \(x\)-half-axis including the origin, i.e., the set

\[ [0, +\infty) \]

Let us verify that, on this restricted domain, the function \(f\) is injective. To this end, we impose \(f(x_1) = f(x_2)\), i.e.,

\[\begin{align*} e^{1+x_1^2} &=e^{1+x_2^2} \\ 1+x_1^2 &=1+x_2^2 \\ x_1^2 &= x_2^2 \Longleftrightarrow x_1=x_2 \end{align*}\]

where the last equivalence holds since \(x \in [0, +\infty)\), i.e., \(x \geq 0\), so the solution \(x_1=-x_2\) is excluded. We have thus proved that the function \(f\) is injective on the restricted domain and, consequently, invertible on \([0, +\infty)\).

We look for the analytic expression of the inverse function \(f^{-1}\) by solving, with respect to the variable \(x\), the equation

\[ e^{1+x^2}=y. \]

The equation is solved by

\[ x=\pm \sqrt{\log{(y)}-1} \]

where, however, it is now clear that the negative solution must be excluded, since \(x\) varies in the restriction \([0, +\infty)\).

Solution

More formally, the inverse function of \(f\) is

\[ f^{-1}(y)=\sqrt{\log{(y)}-1} \]

whose domain is the image of the original function \(f\), that is, \([e, +\infty)\). Hence \(f:[0,+\infty)\to[e, +\infty)\), \(f^{-1}:[e, +\infty) \to [0,+\infty)\).

Exercise 7

Prove that the function

\[ f(x)=\frac{4}{5^{3x}} \]

is invertible on its domain and determine the analytic expression of the inverse \(f^{-1}\). Specify the domain of \(f^{-1}\).

Solution

The domain of the given function is

\[ \left(-\infty,+\infty\right), \]

since the denominator is always nonzero (the exponential never vanishes). To prove that \(f\) is invertible, it suffices to verify that it is injective, i.e.,

\[ \forall x_1,x_2 \in D \qquad f(x_1) = f(x_2) \Longrightarrow x_1 = x_2 \]

Imposing \(f(x_1) = f(x_2)\) is equivalent to imposing that

\[\begin{align*} \frac{4}{5^{3x_1}} &=\frac{4}{5^{3x_2}} \\ 5^{3x_1} &= 5^{3x_2} \\ 3x_1 &= 3x_2 \Longleftrightarrow x_1=x_2 \end{align*}\]

The function \(f\) is injective and, consequently, invertible. We look for the analytic expression of the inverse function \(f^{-1}\) by solving, with respect to the variable \(x\), the equation

\[ \frac{4}{5^{3x}}=y, \]

with \(y >0\), which has the unique solution

\[ x=\frac{1}{3}\log_5{\left(\frac{4}{y}\right)}=-\frac{1}{3}\log_5{\left(\frac{y}{4}\right)} \]

Solution

More formally, the inverse function of \(f\) is

\[ f^{-1}(y)=-\frac{1}{3}\log_5{\left(\frac{y}{4}\right)} \]

whose domain is the image of the original function \(f\), that is, \((0, +\infty)\). Hence \(f:\mathbb{R}\to(0, +\infty)\), while \(f^{-1}:(0, +\infty) \to\mathbb{R}\).

Exercise 8

Prove that the function

\[ f(x)=-1+\sqrt[5]{1+x} \]

is invertible on its domain and determine the analytic expression of the inverse \(f^{-1}\). Specify the domain of \(f^{-1}\).

Solution

The domain of the given function is

\[ \left(-\infty,+\infty\right), \]

since roots of odd index are always defined. To prove that \(f\) is invertible, it suffices to verify that it is injective, i.e.,

\[ \forall x_1,x_2 \in D \qquad f(x_1) = f(x_2) \Longrightarrow x_1 = x_2 \]

Imposing \(f(x_1) = f(x_2)\) is equivalent to imposing that

\[\begin{align*} -1+\sqrt[5]{1+x_1} &=-1+\sqrt[5]{1+x_2} \\ \sqrt[5]{1+x_1} &= \sqrt[5]{1+x_2} \\ 1+x_1 &= 1+x_2 \Longleftrightarrow x_1=x_2 \end{align*}\]

The function \(f\) is injective (it is also strictly increasing) and, consequently, invertible. We look for the analytic expression of the inverse function \(f^{-1}\) by solving, with respect to the variable \(x\), the equation

\[ -1+\sqrt[5]{1+x}=y, \]

which has the unique solution

\[ x=-1+(y+1)^5 \]

Solution

More formally, the inverse function of \(f\) is

\[ f^{-1}(y)=-1+(y+1)^5 \]

whose domain is the image of the original function \(f\), that is, \(\mathbb{R}\). Hence \(f:\mathbb{R}\to\mathbb{R}\), and also \(f^{-1}:\mathbb{R} \to\mathbb{R}\).

Exercise 9

Prove that the function

\[ f(x)=\frac{1-3x}{x+1} \]

is invertible on its domain and determine the analytic expression of the inverse \(f^{-1}\). Specify the domain of \(f^{-1}\).

(Hint: perform the division of \(1-3x\) by \(x+1\)).

Exercise 10

Prove that the function

\[ f(x)=3^{x^2} \]

is invertible on a restriction of its domain and determine the analytic expression of the inverse \(f^{-1}\) on the restriction. Specify the domain of \(f^{-1}\).

Exercise 11

Prove that the function

\[ f(x)=\sqrt{\log{\left(\frac{x-1}{x}\right)}} \]

is invertible on its domain and determine the analytic expression of the inverse \(f^{-1}\). Specify the domain of \(f^{-1}\).