# Proving if a function is One to One

## Homework Statement

Determine whether or not the function is one-to-one and, if so, find the inverse. If the function has an inverse, give the domain of the inverse.
a) f(x) = cos x, x$\in[-pi/2, pi/2]$
b) f(x) = x/|x|
c) f(x) = (2-3x2)3

## The Attempt at a Solution

a) f '(x) = -sinx
When graphing the original function, you can see that it does not pass the horizontal line test. The derivative shows that between this interval, the derivative is not always positive or always negative, it changes between the two which proves that the original function increases and decreases between this interval. Also, f(-pi/3) = f(pi/3) because cos (-pi/3) = 1/2 = cos(pi/3) and that proves that this function is not one-to-one.

b) This is the question i'm most confused about between the three because when i graphed it using wolfram alpha, it is clear that it is not a one to one function because the graph consists of a horizontal line of y = -1 and y = 1 and that clearly will not pass the horizontal line test. However, i just dont know how to show that this function is not one-to-one. This is what i think using what i did before:
1/|1| = 1 = 2/|2| which proves that f(1) = f(2) and that means that this function is not one-to-one

c) I took the derivative of the function:
f'(x) = -18x(2-3x2)2
From the derivative you can see that the function is not always increasing or always decreasing, it changes between the two depending on what the value of x is. I can tell by looking at the function that it has more than one root and if you can show that the function has more than one root that automatically proves that the function is not one-to-one because it will not pass the horizontal line test.

(2-3x2)3 = 0
(2-3x2) = 0
3x2 = 2
x = +-√(2/3)
that means that f(+√(2/3))= f(-√(2/3)) which proves that the function is not one-to-one
Please help me verify if these are correct, and specific help on part b with the absolute value will be greatly appreciated. Thank you.

Part a looks good. You have $cos(-\frac{\pi}{3}) = cos(\frac{\pi}{3})$ but $-\frac{\pi}{3} \neq \frac{\pi}{3}$.