F(x) is differentiable at x =1, find f(x)

1. Nov 5, 2012

Darth Frodo

1. The problem statement, all variables and given/known data

suppose a function f is differentiable at x = 1 and

lim[h → 0] $\frac{f(1 + h)}{h}$ = 5

Finf, f(1) AND f'(1)

3. The attempt at a solution

f(x) is differentiable at x = 1

f(x) is continious at x = 1

lim[x→1] f(x)= f(1)

f'(x) = lim[h→0] $\frac{f(1 + h) - f(1)}{h}$

= 5.lim[h→0]f(1)

= 5.f(1)

= 5lim[x→1]f(x)

Is this correct? I can't see anything else to do.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 5, 2012

clamtrox

No, that's completely wrong :)

$$f'(1) = \lim_{h\rightarrow 0} \frac{f(1+h)-f(1)}{h} = \lim_{h\rightarrow 0} \frac{f(1+h)}{h}-\lim_{h\rightarrow 0} \frac{f(1)}{h}$$

You know what the first term is; it's 5. Now you need to constrain f(1) by the fact that f is differentiable at 1.

3. Nov 5, 2012

HallsofIvy

Staff Emeritus
The denominator goes to 0 so in order that the limit of the fraction exist at all, the numerator must be what?

And $f'(1)= \lim_{h\to 0} (f(1+ h)- f(1))/h$. And now that you know what f(1) is, that limit is precisely what you were given!

Yes, you are correct to here.

I don't see how this follows from the above. You can break the above into
$\lim_{h\to 0} f(1+h)/h- \lim_{h\to 0} f(1)/h= 5- \lim_{h\to 0} f(1)/h$ provided that last limit exists.

No, that's wrong because of what I wrote before.

4. Nov 5, 2012

lurflurf

Consider
f(x)=a+b(x-1)
then generalize