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# Functions: Proof by Induction

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Let n be a natural number, and let f(x) = x^n for all x are members of R.

1) If n is even, then f is strictly increasing hence one-to-one, on [0,infinity) and f([0,infinity)) = [0,infinity).

2) If n is odd, then f is strictly increasing, hence one-to-one, on R and f(R) = R.

"This needs to be a proof by induction, proving the fact that f is strictly increasing. The proof needs to look at the values of n, not x"

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Let n be a natural number, and let f(x) = x^n for all x are members of R.

1) If n is even, then f is strictly increasing hence one-to-one, on [0,infinity) and f([0,infinity)) = [0,infinity).

2) If n is odd, then f is strictly increasing, hence one-to-one, on R and f(R) = R.

"This needs to be a proof by induction, proving the fact that f is strictly increasing. The proof needs to look at the values of n, not x" This is the second time I have posted this problem, the first time someone tried ...

#### Solution Summary

The proof by induction functions are analyzed. A proof is provided as whether a function is strictly increasing.

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