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

## Homework Statement

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)

## 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.

## The Attempt at a Solution

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.

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.

HallsofIvy
Homework Helper

## Homework Statement

suppose a function f is differentiable at x = 1 and

lim[h → 0] $\frac{f(1 + h)}{h}$ = 5
The denominator goes to 0 so in order that the limit of the fraction exist at all, the numerator must be what?

Find, f(1) AND f'(1)
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!

## 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] $\frac{f(1 + h) - f(1)}{h}$
Yes, you are correct to here.

= 5.lim[h→0]f(1)
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.

= 5.f(1)

= 5lim[x→1]f(x)

Is this correct? I can't see anything else to do.
No, that's wrong because of what I wrote before.

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