# 4.2 Activity

## Instructions

(i)Decide if each of the following functions are many-to-one or one-one.
(ii)If a function is many-to-one, give the greatest possible domain over which the function is one-one. (there may be more than one possible answer).

## Notes:

• For each function, assume that $$x \in \mathbb{R}$$.
• There may be more than one correct answer.

Tip: Consider sketching the graph when you are unsure.

(a)$${\rm{a}}(x) = {x^3}$$
(b)$${\rm{b}}(x) = 3x + 5$$
(c)$${\rm{c}}(x) = {x^2} + 2x - 3,\ \ {\rm{ }}x \ge - 1$$
(d)$${\rm{d}}(x) = \sin 2x,{\rm{ }}\ - \frac{\pi }{2} \le x \le \frac{\pi }{2}$$
(e)$${\rm{e}}(x) = \cos x,\ \ {\rm{ }}0 \le x \le \pi$$
(f)\begin{aligned}{\rm{f}}(x) = \frac{1}{{2x - 5}},\ \ {\rm{ }}x \ne 2.5\end{aligned}
(g)$${\rm{g}}(x) = - 2{x^2} + 5x - 2,\,\,{\rm{ }}x \ge 0.5$$
(h)$${\rm{h}}(x) = \sqrt {x - 1} ,\,\,x \ge 1$$
(i)$${\rm{i}}(x) = {x^3} - 4x$$
(j*)\begin{aligned}{\rm{j}}(x) = \frac{1}{{\ln x}}\ \ \end{aligned} (*extension)

(a)
One-to-one
(b)
One-to-one
(c)
One-to-one
(d)
Many-to-one,   $$-\frac{\pi }{4} \le x \le \frac{\pi }{4}$$
(e)
One-to-one
(f)
One-to-one
(g)
Many-to-one,   $$x \ge \frac{5}{4}$$ (alternate answer: $$x \le \frac{5}{4}$$)
(h)
One-to-one
(i)
Many-to-one,   $$x \ge \frac{2}{{\sqrt 3 }}$$
(j)
One-to-one