Mean theorim question

1. Nov 21, 2008

transgalactic

2. Nov 21, 2008

Unco

If h(x) = f(x) for all x then h'(x) = f'(x).

Last edited: Nov 21, 2008
3. Nov 21, 2008

transgalactic

for a fact
??

4. Nov 21, 2008

Unco

Yes; consider the definition $$h'(x) = \lim_{t\to 0}\frac{h(x+t)-h(x)}{t}$$.

5. Nov 21, 2008

Pere Callahan

If f(x)=g(x) for all x, then the two functions are the same. And for this very reason they have the same derivative.

6. Nov 21, 2008

transgalactic

7. Nov 21, 2008

Unco

Now you're just confused by notation. Let's use some neutral letters.

If A and B are differentiable functions on [a,b] such that A(x) = B(x) for all x in [a,b], then A'(x) = B'(x).

Why? $$A'(x) = \lim_{t\to 0}\frac{A(x+t)-A(x)}{t} = \lim_{t\to 0}\frac{B(x+t)-B(x)}{t} = B'(x)$$.

Also, they said "h'(c)=f'(c)-g'(c)" NOT "h'(c)=f'(x)-g'(x)".

8. Nov 21, 2008

transgalactic

when you say "A(x) = B(x) for all x in [a,b]"

that mean that they share each one of their points
which means that A(x) and B(x) are the same function then of course A'(x) = B'(x) (its the same function)

which is not true in the graph

Last edited: Nov 21, 2008
9. Nov 21, 2008

Pere Callahan

Take A(x)=h(x) and B(x)=f(x)-g(x)

10. Nov 21, 2008

ahhhh thanks