Rolle's Theorem differentiation

[Fairy111]

Homework Statement

Let f be continuous and differentiable on [a,b], and suppose that f attains its maximum and minimum points c and d, respectively, where c,d belong to [a,b]. Show that f ' (d) = 0

Homework Equations

The Attempt at a Solution

I thought about using the Mean Value Theorem which states that:

f ' (d)=f(b) - f(a) / (b-a)

but then didn't know how to continue.

Or maybe use Rolle's Theorem, but i couldn't see how that would show that f ' (d) = 0

Any help would be great thanks.

 

[LogicalTime]

this isn't necessarily true
$f(x) = x^2$
$x \in [0,10]$
is continuous and differentiable, f attains it's maximum at f(10)=100 but f'(10) =20. I think you need c,d to belong to (a,b). This way you know that f(b)< f(d) and f(a)< f(d)

 

[kmeado07]

c,d do belong to [a,b]

so f(a)<f(d), but f(b)>f(d)

I don't see how this helps me to show that f ' (d) = 0 thought.

 

[HallsofIvy]

No, the mean value theorem doesn't tell you that: it says that (f(b)- f(a))/(b-a)= f'(h) for SOME number h between a and d. It does not follow that h is the "d" where this is minimum.

Rather, use the basic definition of the derivative.
$$f'(d)= \lim_{h\rightarrow 0} \frac{f(d+h)- f(d)}{h}$$

If d is a (local) minimum, what can you say about f(d+ h) for any small non-zero h?

 

[kmeado07]

Is the limit when h tends to 0?...then it would be f(d)-f(d)/h, so f '(d) would equal zero.

Im not sure what f(d+h) implies though...

 

[LogicalTime]

first things first
$c,d \in (a,b)$
instead of [a,b] if you need f'(x) = 0 at min/max points
agreed?

 

[Fairy111]

f(c) is the maximum and f(d) is the minimum.

What can i say about f(d+h) ?

 

[gabbagabbahey]

f(d) is the minimum, sooo f(d+h) is...

(1) smaller than f(d)
(2) greater than f(d)
(3)purple

?

 

[Fairy111]

ok...so obviously f(d+h)>f(d)

i still don't see how this can help me show that f '(d) = 0

 

[HallsofIvy]

Is the limit when h tends to 0?...then it would be f(d)-f(d)/h, so f '(d) would equal zero.

Im not sure what f(d+h) implies though...

No, that does not follow. Every derivative becomes "0/0"!

 

[HallsofIvy]

ok...so obviously f(d+h)>f(d)

i still don't see how this can help me show that f '(d) = 0

So f(d+h)- f(d)> 0 no matter whether h is positive or negative.

If h> 0, what can you say about (f(d+h)- f(d))/h?

If h< 0, what can you say about (f(d+h)- f(d))/h?

And so what must be true of
$$\lim_{h\rightarrow 0} \frac{f(d+h)- f(d)}{h}$$?

 

[Fairy111]

ok, so if h>0, then (f(d+h)- f(d))/h >0

and if h< 0, then (f(d+h)- f(d))/h <0

which means that the limit eqn will be zero.

 

[HallsofIvy]

WHY does that mean the limit must be 0?

 

[Fairy111]

i don't know, I am sorry, i just don't understand where you're going with this...

 

[gabbagabbahey]

If $$\lim_{x \to 0}$$ of some expression exists, what must be true about $$\lim_{x \to 0^+}$$ and $$\lim_{x \to 0^-}$$?

In this case, you are told that f is differentiable, so you know that the limit $$\lim_{h\rightarrow 0} \frac{f(d+h)- f(d)}{h}$$ must exist. So...