# Spivak Derivate Question

1. Nov 11, 2006

### redyelloworange

Suppose
f(a) = g(a) = h(a)
and f(x) <= g(x) <= (x) for all x

Prove g(x) is differentible and that
f'(a) = g'(a) = h'(a).

So.. I need to prove that the following limit exists:

lim h -->0 (g(x+h) - g(x)) / h

but how can i use the fact that f(x) <= g(x) <= (x) for all x?

Thanks

2. Nov 11, 2006

### StatusX

Obviously you're gonna have to tell us something about f and h.

3. Nov 11, 2006

### redyelloworange

f(a) = g(a) = h(a)
and f(x) <= g(x) <= h(x) for all x

and

f'(a) = h'(a)

Last edited: Nov 11, 2006
4. Nov 11, 2006

### HallsofIvy

Staff Emeritus
Is it possible that f and h are given as differentiable? Otherwise, just take f(x)= g(x)= h(x) to be any non-differentiable functions and the hypotheses are satisfied while the conclusion is not true!

5. Nov 11, 2006

### redyelloworange

you're right

*I seem to have typed the question wrong many times*
sorry.

So, given:

f(a) = g(a) = h(a)
and
f(x) <= g(x) <= h(x) for all x
and
f'(a) = h'(a)

Prove g is differentiable at a, and that g'(a) = g'(a) = h'(a).

6. Nov 11, 2006

### StatusX

Do you know the squeeze theorem?

7. Nov 11, 2006

### redyelloworange

oh!
sol instead of just including g(x) in the inequality.. i include more:

f(x) <= g(x) <= h(x)
f(a+h) <= g(a+h) <= h(a+h)
f(a+h) - f(a) <= g(a+h) - f(a) <= h(a+h) - f(a)
f(a+h) - f(a) <= g(a+h) - f(a) <= h(a+h) - f(a)

f'(a) <= g'(a) <= h'(a)

but f'(a) = h'(a)

so (how exactly do i phrase this?)

f'(a) = g'(a) = h'(a)

ooh. that was a good hint/spark =) thanks!