Worked examples · Sheet 3

Inverse functions and composite functions

Theory used: Inverse function Composite function Injective function Exponential function Logarithm Vertex of the parabola

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Find the inverse of $f(x) = 2e^{x} + 1$ , with its domain, and verify that $f^{-1}(f(x)) = x$ .

y = 2e^{x} + 1 \iff e^{x} = \frac{y - 1}{2} \iff x = \ln\frac{y - 1}{2}, \qquad y > 1

f^{-1}(y) = \ln\frac{y - 1}{2}, \qquad f^{-1} : (1, +\infty) \to \mathbb{R}

f^{-1}(f(x)) = \ln\frac{2e^{x} + 1 - 1}{2} = \ln e^{x} = x

The function $e^{x}$ is increasing, so $2e^{x} + 1$ is too, and it is therefore injective on $\mathbb{R}$ . To find the inverse, solve the equation $y = f(x)$ for $x$ : isolate the exponential and then take the natural logarithm, which is the inverse of $e^{x}$ . The logarithm requires a positive argument, therefore $\dfrac{y - 1}{2} > 0$ , that is, $y > 1$ . The domain of the inverse is the range of $f$ , and indeed $f(x) = 2e^{x} + 1$ takes exactly the values greater than $1$ , since $e^{x}$ takes all positive values. The check confirms that $f^{-1}$ undoes what $f$ does.

Exercise 2

Let $f(x) = \sqrt{x}$ and $g(x) = x - 4$ . Write the composite functions $g \circ f$ and $f \circ g$ and find their domains.

(g \circ f)(x) = g(\sqrt{x}) = \sqrt{x} - 4, \qquad D_{g \circ f} = [0, +\infty)

(f \circ g)(x) = f(x - 4) = \sqrt{x - 4}, \qquad x - 4 \geq 0 \iff x \geq 4, \qquad D_{f \circ g} = [4, +\infty)

In the composite function $g \circ f$ , first $f$ is applied and then $g$ . For $g \circ f$ we first compute $\sqrt{x}$ , which requires $x \geq 0$ , and then subtract $4$ , an operation that imposes no conditions; the domain is therefore $[0, +\infty)$ . For $f \circ g$ we first compute $x - 4$ and then take its square root, which requires $x - 4 \geq 0$ ; the domain is $[4, +\infty)$ . The two compositions are different, both in the formula and in the domain; for example, for $x = 9$ the first equals $3 - 4 = -1$ and the second $\sqrt{5}$ .

Exercise 3

Find the inverse of the function $f(x) = x^2 - 4x + 1$ restricted to the interval $[2, +\infty)$ , with its domain.

f(x) = (x - 2)^2 - 3, \qquad x \geq 2

y = (x - 2)^2 - 3 \iff (x - 2)^2 = y + 3 \iff x - 2 = \sqrt{y + 3}

f^{-1}(y) = 2 + \sqrt{y + 3}, \qquad f^{-1} : [-3, +\infty) \to [2, +\infty)

On all of $\mathbb{R}$ the function is not injective, because the parabola takes every value greater than the minimum twice. Completing the square shows that the vertex is $(2, -3)$ ; to the right of the vertex the function is increasing, so its restriction to $[2, +\infty)$ is injective and has range $[-3, +\infty)$ . To find the inverse, solve $y = (x - 2)^2 - 3$ for $x$ . When taking the square root, choose the positive sign, because $x \geq 2$ implies $x - 2 \geq 0$ . The inverse is $f^{-1}(y) = 2 + \sqrt{y + 3}$ , defined for $y \geq -3$ ; as a check, $f(3) = -2$ and $f^{-1}(-2) = 2 + 1 = 3$ .