# Mean Value Theorm type question

playboy

## Homework Statement

Let f : R ---> R be a Continuosly differentiable function such that f(0) = 0 , f(1) = 1 and f'(0) = f'(1) = 0 .
Show that their exists an S in (0, 1) such that |f''(S)| > 4 .

## Homework Equations

I am unsure if I am even supposed to use the mean value theorm in this problem, but it seems like you have to..

THE MEAN VALUE THEORM:

Let f : [a, b] → R be continuous on the closed interval [a, b], and differentiable on the open interval (a, b). Then there exists some R in (a, b) such that

f(b) − f(a) = f'(R)(b − a)

## The Attempt at a Solution

Apply derivatives to the mean value theorm above to get:

f'(b) − f'(a) = f''(S)(b − a)

Now, taking the magnitudes, we get

|f'(b) − f'(a)| < |f''(S)||(b − a)|

Now computing, we get

|f'(1) − f'(0)| < |f''(S)||(1 − 0)|

|0 − 0| < |f''(S)|

0< |f''(S)|

Hence

0< |f''(S)|

so indeed, 4< |f''(S)|

How does this sound?

I am very unsure of this...I dont think it is rigourous enough

HallsofIvy
Homework Helper
You are only given that f is "continuously differentiable". It does not follow from that that f even has a second derivative at any particular point.

Also, f(b)- f(a) = f'(c)(b-a) is true for specific values of a, b, c, not variables. You can say that, for any x, y in [a, b], f(x)- f(y)= f'(c)(x-y) but then what variable are you going to differentiate with respect to: x or y?

Finally, you can say that, for any x in [a,b], f(x)- f(b)= f'(c)(x-a) and differentiate that with respect to x but then you get f'(x)= f'(c), not what you give.

You can use the mean value theorem to say that there exist c in [0, 1] such that f'(c)= f(1)- f(0)= 1.

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playboy
You are only given that f is "continuously differentiable". It does not follow from that that f even has a second derivative at any particular point.

Also, f(b)- f(a) = f'(c)(b-a) is true for specific values of a, b, c, not variables. You can say that, for any x, y in [a, b], f(x)- f(y)= f'(c)(x-y) but then what variable are you going to differentiate with respect to: x or y?

Finally, you can say that, for any x in [a,b], f(x)- f(b)= f'(c)(x-a) and differentiate that with respect to x but then you get f'(x)= f'(c), not what you give.

You can use the mean value theorem to say that there exist c in [0, 1] such that f'(c)= f(1)- f(0)= 1.

Thank you HallsofIvy... I had the Mean Value Theorm wrong...

Let me start again..

a=0, b=1

for any x in [a,b], f(x)- f(b)= f'(c)(x-a)

hence,

f(x)- f(1)= f'(c)(x-0)

and we take the derivative with respect to x

f'(x)- f'(1)= f''(c)(x-0)

now we take the magnitude

|f'(x)- f'(1)| < |f''(c)||x|

and f'(1)=0

|f'(x)| < |f''(c)||x|

and now i don't know where to go from here..

playboy
I think i know how to solve it...

I need to find a function such that f(0)=0 and f(1)=1 and f'(0)=f'(1)=1.

does:

f(x) = 0 for 0 =< x =< 0.25

f(x) = x for 0.25 < x =<0.75

f(x) = 1 for 0.75< x =< 1

but off course, f'(x) wont be conintuously differentiable...

thus, for

|f'(x)| < |f''(c)||x|

choose x=0.75

hence,

for x=0.75, f'(x) = 1

1<|f''(c)||0.75|
4/3 < |f''(c)|

but this function dosn't work in my case...

anybody know a few functions such that f(0)=0 and f(1)=1 and f'(0)=f'(1)=1 ?