1. ## Differentiation

I need to prove from the definition of defferntiability that the function

f(x)= 2x-1/x-2

is differentiable at the point 1 and find f'(1)

So far I have

The difference quotient for f at 1 is

Q(h) = f(h)-f(1)/h

= 2h-1/h-2 + 1 / h

= ?

can anyone help?

2. ## Re: Differentiation

When is a function differentiable? ...

3. ## Re: Differentiation

find a common denominator, cancel any like terms and then find the limit of h->0, if you get (some number)/0 then the function is not differentiable, however if you get 0/0 you get an indeterminate so then you must try another technique (the second derivative).

4. ## Re: Differentiation

I'm sorry, maybe my question was unclear, but you have to calculate (like abdulsafan said). A function is differentiable in a point of her domain if she has a derivative.

Notice you have to calculate
lim_(h->0) (f(1+h)-f(1))/h

5. ## Re: Differentiation

would that = h/1+h?

6. ## Re: Differentiation

You have to calculate:
$\lim_{h \to 0}\frac{f(1+h)-f(1)}{h}$

and
$f(1+h)=\frac{2(1+h)-1}{(1+h)-2}=\frac{2h+1}{h-1}$
$f(1)=-1$
so
$\lim_{h \to 0} \frac{\frac{2h+1}{h-1}+1}{h}=\lim_{h \to 0}\frac{\frac{2h+1+(h-1)}{h-1}}{h}$
$=\lim_{h \to 0} \frac{3h}{(h-1)h}=-3$

Therefore $f'(1)=-3$

Check:
$D\left(\frac{2x-1}{x-2}\right)=\frac{D(2x-1)\cdot (x-2)-(2x-1)\cdot D(x-2)}{(x-2)^2}=\frac{2\cdot (x-2)-(2x-1)}{(x-2)^2}=\frac{-3}{(x-2)^2}$
$f'(1)=\frac{-3}{(1-2)^2}=-3$

Therefore $f'(1)=-3$

Do you understand now? ...

7. ## Re: Differentiation

Sorry I should have said this before: a function is differentiable if
A) f(a) exist
B) limit as x approaches a exist
C) f'(a) exist
D) and f(a) equals the limit as x approaches a

I think that ur function does all of the above...hope this Helps!

8. ## Re: Differentiation

definition of a derivative at a point ...

$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$

if it exists ...

$f'(1) = \lim_{x \to 1} \frac{f(x) - f(1)}{x-1}$

$f'(1) = \lim_{x \to 1} \frac{\frac{2x-1}{x-2} + 1}{x-1}$

$f'(1) = \lim_{x \to 1} \frac{(2x-1) + (x-2)}{(x-1)(x-2)}$

$f'(1) = \lim_{x \to 1} \frac{3x-3}{(x-1)(x-2)}$

$f'(1) = \lim_{x \to 1} \frac{3(x-1)}{(x-1)(x-2)}$

$f'(1) = \lim_{x \to 1} \frac{3}{x-2} = -3$