Proving f(x)=k on [a,b] using Rolle's theorem and other calculus theorems

[Zarem]

Homework Statement

Hey, I'm revising my calculus 1 test, I'm just not sure if this is a valid way to prove this.

If f is continuous on [a,b] and if f'(x)=0 for every x in (a,b), prove that f(x)=k for some real number k

Homework Equations

Rolle's theorem: if f(a)=f(b), then there is some number c such that f'(c)=0

Possibly other theorems like MVT, IVT whatever

The Attempt at a Solution

Since f is continuous on [a,b] and differentiable on (a,b), Rolle's theorem applies.

If f'(x)=0 for every x in (a,b), then there is a number Z in the interval [c,d], where a[tex]\leq[/tex]c[tex]\leq[/tex]d[tex]\leq[/tex]b, such that (f(d)-f(c))/(d-c)=f'(Z)

Since Z is in [a,b] f'(Z)=0, so for any interval [c,d] (f(d)-f(c))/(d-c)=0, therefore f(c)=f(d) which means f(x)=k for some real number k

 

[gb7nash]

I would approach this a different way through a contradiction proof. Suppose f(x) is not constant on [a,b]. So there exists c,d in [a,b] such that f(c) does not equal f(d). Do you see where you can apply mean value theorem? (assuming you can use MVT)

 

[Zarem]

I would approach this a different way through a contradiction proof. Suppose f(x) is not constant on [a,b]. So there exists c,d in [a,b] such that f(c) does not equal f(d). Do you see where you can apply mean value theorem? (assuming you can use MVT)

If f(c)=/= f(d) then by the MVT there exists a f'(x) in [a,b] that does not equal zero.

Thanks, but do you think the way I did it is also correct?

 

[Zarem]

Anyone else know if the first way I did it (or a similar way) is correct, I only ask because I think that's how my professor wants it done.

 

[gb7nash]

Since f is continuous on [a,b] and differentiable on (a,b), Rolle's theorem applies.

Omit this. If you look back at your answer, Rolle's Theorem is never used, since we're not assuming that f(c) = f(d). That's the conclusion we want to arrive at.

If f'(x)=0 for every x in (a,b), then there is a number Z in the interval [c,d], where a[tex]\leq[/tex]c[tex]\leq[/tex]d[tex]\leq[/tex]b, such that (f(d)-f(c))/(d-c)=f'(Z)

It looks like you have the right idea on this, though you might want to rearrange your sentence. One way of showing that a function is constant is by fixing two arbitary points not equal to each other (c and d in this case) and arriving at [tex]f(c) = f(d)[/tex]. Here might be a better way to reword it:

Fix c and d in [a,b] such that [tex]a \leq c < d \leq b[/tex]. (Notice that we want to choose two arbitrary points, so WLOG c < d). Since f'(x) = 0 for all x in (c,d), f is differentiable on (c,d), so by MVT there exists a Z in (c,d) such that (f(d)-f(c))/(d-c)=f'(Z)

We know f'(Z) = 0, so:
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