Worked examples · Sheet 2

Range, monotonicity and injectivity

Theory used: Real function of a real variable Increasing and decreasing functions Injective function Vertex of the parabola

Go to: Exercise 1 Exercise 2

Exercise 1

Let $f(x) = x^2 - 6x + 5$ be defined on the interval $[0, 5]$ . Find the range of $f$ , determine whether $f$ is injective and find the largest interval containing $5$ on which $f$ is injective.

x_V = -\frac{-6}{2} = 3, \qquad f(3) = 9 - 18 + 5 = -4

f(0) = 5, \qquad f(5) = 25 - 30 + 5 = 0

f([0, 5]) = [-4, 5]

f(1) = 1 - 6 + 5 = 0 = f(5)

y = x^{2} - 6 x + 5

[0, 5]

the parabola falls from

5

to the minimum

- 4

x = 3

and rises back to

0

: the range is

[- 4, 5]

, and the value

0

is taken twice

The function is quadratic with $a = 1 > 0$ , so by the theorem on the vertex of the parabola its minimum value is attained at $x_V = 3$ and equals $-4$ . The vertex lies in the interval $[0, 5]$ ; to the left of the vertex the function decreases and to the right it increases, therefore the maximum value on the interval is the larger of the values at the endpoints, $f(0) = 5$ and $f(5) = 0$ . The graph on $[0, 5]$ is an unbroken arc of a parabola, so the function takes all the values between the minimum and the maximum, and the range is $[-4, 5]$ ; this property will be justified by the intermediate value theorem in the topic on limits and continuity.

The function is not injective on $[0, 5]$ , because the two different values $1$ and $5$ have the same image $0$ . It is instead increasing, and hence injective, on the interval $[3, 5]$ , which lies to the right of the vertex. No larger interval containing $5$ works, because adding points to the left of $3$ gives values already taken to the right of the vertex.

Exercise 2

Let $f(x) = 3 - \dfrac{2}{x}$ for $x > 0$ . Prove that $f$ is increasing and find its range.

0 < x_1 < x_2 \implies \frac{1}{x_1} > \frac{1}{x_2} \implies -\frac{2}{x_1} < -\frac{2}{x_2} \implies f(x_1) < f(x_2)

f(x) = y \iff \frac{2}{x} = 3 - y \iff x = \frac{2}{3 - y}, \qquad x > 0 \iff y < 3

f\big((0, +\infty)\big) = (-\infty, 3)

By the definition of increasing function, we compare the images of any two values $x_1 < x_2$ of the domain. For positive numbers the reciprocal reverses inequalities, so $\dfrac{1}{x_1} > \dfrac{1}{x_2}$ ; multiplying by $-2$ , which is negative, the inequality is reversed again, and adding $3$ to both sides gives $f(x_1) < f(x_2)$ . Therefore $f$ is increasing on $(0, +\infty)$ .

To find the range, we look for the values $y$ for which the equation $f(x) = y$ has a solution in the domain. Isolating $x$ gives $x = \dfrac{2}{3 - y}$ , which makes sense for $y \neq 3$ and is positive exactly when $3 - y > 0$ . Hence a value $y$ is taken if and only if $y < 3$ , and the range is $(-\infty, 3)$ . The value $3$ is never reached, because $\dfrac{2}{x}$ is always positive.