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So, yesterday in class we were asked to try and solve the following problem:

Given a, b [itex]\in[/itex] R with a < b, draw the graph of an example of a continuous function f such that f: [a,b] [itex]\rightarrow[/itex] R, f(a) = f(b), and there does not exist c [itex]\in[/itex] (a, b) such that f'(c) = 0.

Now in class we arrived at the conclusion that there exists no such solution, however I beg to disagree.

Let's suppose that a = -1 and b = 1 which satisfies the inequality. Now for f(a) to = f(b) one function comes to mind, the associated power series:

f(x) = 1 + 2x + 4x

Such that, f

= [itex]\frac{d}{dx}[/itex]([itex]\frac{1}{1-2x}[/itex])

= ([itex]\frac{-4}{2x - 1}[/itex])

Therefore, there does not exist f'(c) = 0

If we take the limit as x [itex]\rightarrow[/itex] [itex]\infty[/itex] then f'(x) = 0 however there is no particular element c of (a, b) s.t. f'(c) = 0.

My question is am I right? If not where did I go horribly wrong.

Given a, b [itex]\in[/itex] R with a < b, draw the graph of an example of a continuous function f such that f: [a,b] [itex]\rightarrow[/itex] R, f(a) = f(b), and there does not exist c [itex]\in[/itex] (a, b) such that f'(c) = 0.

Now in class we arrived at the conclusion that there exists no such solution, however I beg to disagree.

Let's suppose that a = -1 and b = 1 which satisfies the inequality. Now for f(a) to = f(b) one function comes to mind, the associated power series:

f(x) = 1 + 2x + 4x

^{2}+ 8x^{3}+ ... + 2^{n}x^{n}= [itex]\frac{1}{1-2x}[/itex]Such that, f

^{2}(1) = 1 and f^{2}(-1) = 1 therefore, f(a) = f(b).= [itex]\frac{d}{dx}[/itex]([itex]\frac{1}{1-2x}[/itex])

^{2}= ([itex]\frac{-4}{2x - 1}[/itex])

^{3}Therefore, there does not exist f'(c) = 0

If we take the limit as x [itex]\rightarrow[/itex] [itex]\infty[/itex] then f'(x) = 0 however there is no particular element c of (a, b) s.t. f'(c) = 0.

My question is am I right? If not where did I go horribly wrong.

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