Proving 1-1 Functions with f(a) = f(b) and a = b: x^3-2 and x^4+2

[jwxie]

Homework Statement

[1] f(x) = x^3-2
[2] f(x) = x^4+2

Homework Equations

f(a) = f(b)
a = b

The Attempt at a Solution

Assuming that f(a) = f(b), for which some a =/ b

So for [1] f(x) = x^3-2, let assume a=/b, that f(a) = f(b)
in the end I got, a^3 = b^3

for [2], I did the same thing, and for f(x) = x^4+2, in the end I got a^4 = b^4

Now I need to simply them down to a = b, so I think I need to take cubic root on both side, and square root on both side, respectively.

Now why is [2] not a 1-1 function, while [1] is a 1-1 function?
When I solve for x, for example, in the same of y = x^2, i get sqrt of y = x, so it is not a 1-1 function.

What about x^3-2? I got cub root of y + 2 = x
Thanks

 

[Mark44]

Homework Statement

[1] f(x) = x^3-2
[2] f(x) = x^4+2

Homework Equations

f(a) = f(b)
a = b

The Attempt at a Solution

Assuming that f(a) = f(b), for which some a =/ b

So for [1] f(x) = x^3-2, let assume a=/b, that f(a) = f(b)
in the end I got, a^3 = b^3

for [2], I did the same thing, and for f(x) = x^4+2, in the end I got a^4 = b^4

Now I need to simply them down to a = b, so I think I need to take cubic root on both side, and square root on both side, respectively.

The first one is usually called the cube root. For the other you need a fourth root, not a square root.

You don't need the cube root and fourth root, though. In fact, it's best to not use them. For the first problem you have
a³ = b³ <==> a³ - b³ = 0.
The last equation can be factored to give three solutions, two of which are complex.

For the other problem, you have
a⁴ = b⁴ <==> a⁴ - b⁴ = 0.
This equation can be factored to give four solutions, two of which are complex.

You can ignore the complex solutions. Can you find the real solutions to these equations?

Now why is [2] not a 1-1 function, while [1] is a 1-1 function?
I mean sometime cubic root can have -1, -1, 1..

Thanks

 

[jwxie]

Hi, thanks.

a3 = b3 <==> a3 - b3 = 0.
The last equation can be factored to give three solutions, two of which are complex.

For the other problem, you have
a4 = b4 <==> a4 - b4 = 0.
This equation can be factored to give four solutions, two of which are complex.

What I do not understand is how can I tell

When I solve for x, for example, in the same of y = x^2, i get sqrt of y = x, so it is not a 1-1 function.

What about x^3-2? I got cub root of y + 2 = x

For your question, I didn't really think about those complex number (since I didn't even know there are complex numbers in those roots).
I guess, from your information, you just literally pointed out that [1] there is only one a (b), and [2] there is two a, b that can still produce 0 (the value of y).

 

[Mark44]

Can you solve the equation a⁴ - b⁴ = 0 by factoring? This is the same as ((a²)² - (b²)² = 0.

Can you solve the equation a³ - b³ = 0 by factoring?

 

[HallsofIvy]

$(-1)^3\no 1^3$

$(-1)^4 = 1^5$