# Thread: Determine whether the following functions are bijections

1. ## Determine whether the following functions are bijections

Determine whether or not the following functions from real numbers to real numbers are bijections (I don't need help with this part), and explain why or why not (I need help with THIS part).

a) $\displaystyle f(x)=-4x+5$ (Bijective)
b) $\displaystyle f(x)=2x^3-11$ (Bijective)
c) $\displaystyle f(x)=3x+2$ (Bijective)
d) $\displaystyle f(x)=|4x^3-5|$ (Not bijective)
e) $\displaystyle f(x)=2x^4-5$ (Not bijective)

The main issue I'm having with this problem is proof presentation. I don't exactly know how to explain whether or not a function is bijective or not.

Of course, if I've made a mistake, I'd appreciate it if you let me know.

2. Here is some good news: All linear functions, $\displaystyle y=f(x)=mx+b,~m\ne 0$ are bijections.
In part b) look at the derivative.
For the others, find counter examples.

3. Dear Runty,

I have done the first one for you. All the other problems could be solved using this method.

$\displaystyle f(x)=-4x+5$

If $\displaystyle a,b\in{R}~such~that~f(a)=f(b)$

$\displaystyle -4a+5=-4b+5$

$\displaystyle a=b$

Therefore, $\displaystyle f(a)=f(b)\Rightarrow{a=b}~\forall~{a,b\in{R}}$

Hence $\displaystyle f$ is injective.

Suppose, $\displaystyle a\in{Ran(f)}$

$\displaystyle If~ b=\frac{a-5}{-4}$

Since, $\displaystyle a\in{R}\Rightarrow{b=\frac{a-5}{-4}\in{R}}$

Therefore, $\displaystyle f(b)=f\left(\frac{a-5}{-4}\right)=a$

$\displaystyle f(b)=a$

$\displaystyle \exists~{b\in{R}}~\forall~{a\in{Ran(f)}}$

Hence $\displaystyle f$ is surjective.

Therefore f is bijective.

4. Thank you both for your answers. Looking at this again, however, I'm wondering how I'd prove how the fourth one is not bijective, mainly since it's an absolute function.

$\displaystyle f(x)=|4x^3-5|$

I actually can't find a situation where $\displaystyle f(x)=f(y)$. Anyone have a suggestion so I can prove that it's not injective, and therefore not bijective?

5. Originally Posted by Runty
Thank you both for your answers. Looking at this again, however, I'm wondering how I'd prove how the fourth one is not bijective, mainly since it's an absolute function.

$\displaystyle f(x)=|4x^3-5|$

I actually can't find a situation where $\displaystyle f(x)=f(y)$. Anyone have a suggestion so I can prove that it's not injective, and therefore not bijective?
Example:
Take $\displaystyle f(x)=1 \implies 4x^3-5=1 \implies x=\sqrt[3]{\frac{6}{4}}$ And the other case is that the argument of the absolute value gives you -1: $\displaystyle 5-4x^3=1 \implies x=1$. So, there you have two different x's with the same f(x)

6. Originally Posted by felper
Example:
Take $\displaystyle f(x)=1 \implies 4x^3-5=1 \implies x=\sqrt[3]{\frac{6}{4}}$ And the other case is that the argument of the absolute value gives you -1: $\displaystyle 5-4x^3=1 \implies x=1$. So, there you have two x's with the same y
EDIT: Nevermind, I misread your answer. Thanks for the help.

7. hey runty,
maybe it helps to think about the mapping:

$\displaystyle x \mapsto \Vert 4x^3 - 5 \Vert$

which is the same as:

$\displaystyle x \mapsto 4x^3 - 5$ or $\displaystyle x \mapsto -(4x^3 - 5)$

the question is are there two elements from x (domain) which map to the same value:

$\displaystyle 4x_1^3 - 5 = -(4x_2^3 - 5)$

8. Originally Posted by Runty
Are you sure about this? I'm pretty sure this function is not bijective.
That's the counter example to show that the function is not bijective, man. The bijective functions asociates each f(x) with a unique x. In this case, we have a f(x) asociated to two x. Then, the function is not bijective.