Having the function fn(x)=x^n+n*x-1, demonstrate that fn is convex, for any n>2; n is a natural number.

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A function is convex over a region R if every point on the function in R lies beneath the line connecting the boundaries of R. The second derivative of a function can tell us whether a function is convex over a region: if and only if f''(x) > 0, then...

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A function is convex over a region R if every point on the function in R lies beneath the line connecting the boundaries of R. The second derivative of a function can tell us whether a function is convex over a region: if and only if f''(x) > 0, then the region is convex.

So for f(x) = x^n + nx - 1

f'(x) = nx^(n-1) + n

f''(x) = n(n-1)x^(n-2)

Note that the natural numbers are the set {0, 1, 2, 3, ...}

So for n = {0,1}, f''(x) = 0

For n = 2, f''(x) = 2, so f(x) is convex.

For n > 2, f''(x) = N*x^(n-2), where N > 0 and n-2 > 0

Hence, for n-2 even f(x) is strictly convex for all x, and for n-2 odd f(x) is convex for x > 0.

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