Understanding the Mean Theorem: Evaluating h'(c) = f'(c) - g'(c)

[transgalactic]

i know that h(x)=f(x)-g(x)
but on what basis they conclude that

h'(c)=f'(c)-g'(c)


http://img356.imageshack.us/img356/5427/69700925ii7.gif

??

 

[Unco]

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

 

[transgalactic]

for a fact
??

 

[Unco]

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

 

[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.

 

[transgalactic]

but f(x) is a curve and g(x) is a line
f(x) is not g(x)
and so is h(x) differs them all
look at the link

http://img356.imageshack.us/img356/5427/69700925ii7.gif

all the lines are different

h(c)=f(x)-g(x)

how they came to
h'(c)=f'(x)-g'(x)

 

[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)".

 

[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

 

[Pere Callahan]

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

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

 

[transgalactic]

ahhhh thanks