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

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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 ...

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- BSc , Wuhan Univ. China
- MA, Shandong Univ.

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- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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