An Impossible Solution: Solving the Function Graphing Problem in Calculus Class

[JPBenowitz]

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

Given a, b $\in$ R with a < b, draw the graph of an example of a continuous function f such that f: [a,b] $\rightarrow$ R, f(a) = f(b), and there does not exist c $\in$ (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 + 4x² + 8x³ + ... + 2ⁿxⁿ = $\frac{1}{1-2x}$

Such that, f²(1) = 1 and f²(-1) = 1 therefore, f(a) = f(b).

= $\frac{d}{dx}$($\frac{1}{1-2x}$)²

= ($\frac{-4}{2x - 1}$)³

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

If we take the limit as x $\rightarrow$ $\infty$ 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.

 

[paulfr]

Going too fast ?
Differentiation that power series and see that at x=0
f ' (x) = 0

 

[JPBenowitz]

Going too fast ?
Differentiation that power series and see that at x=0
f ' (x) = 0

I accidently posted before I finished lol.

 

[gb7nash]

This statement is indeed true.

I'm not sure why you're taking a limit as x goes to infinity. In any case, if you think the statement is wrong, you need to show there exists a c in (a,b) such that f'(c) = 0 for every f(x). Unfortunately, showing one example doesn't prove anything.

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

Consider the example f(x) = |x| and a = -1, b = 1. Does this satisfy the hypothesis? Is the conclusion satisfied?

 

[JPBenowitz]

This statement is indeed true.

I'm not sure why you're taking a limit as x goes to infinity. In any case, if you think the statement is wrong, you need to show there exists a c in (a,b) such that f'(c) = 0 for every f(x). Unfortunately, showing one example doesn't prove anything.



Consider the example f(x) = |x| and a = -1, b = 1. Does this satisfy the hypothesis? Is the conclusion satisfied?

Which statement is true? That there is no solution? Or that there is a solution?

 

[gb7nash]

This statement:

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

It is possible to construct an example of a continuous function f such that there does not exist a c $\in$ (a,b) such that f'(c) = 0. Consider f(x) = |x|.

 

[JPBenowitz]

This statement:



It is possible to construct an example of a continuous function f such that there does not exist a c $\in$ (a,b) such that f'(c) = 0. Consider f(x) = |x|.

f(x) is discontinuous at 0.

 

[gb7nash]

f(x) is discontinuous at 0.

f(x) = |x| is continuous at x=0. However, it isn't differentiable at x=0.

 

[JPBenowitz]

f(x) = |x| is continuous at x=0. However, it isn't differentiable at x=0.

Sorry, that's what I meant.

 

[AlephZero]

In your OP, f(x) = 1/(1-2x) is not continuous (in fact it is not even defined) when x = 1/2.