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jwxie

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## 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

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