Question

If f is differentiable at 1, and the limit of f(1+h)/h as h->0 is 5, what is f(1)?

Answer 1

@jim_thompson5910

Answer 2

` f is differentiable at 1`
so \[\Large \lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\] exists and it is defined based on the limit definition of the derivative

Answer 3

so \[f'(1) = 5-\lim_{h \rightarrow 0}\frac{f(1)}{h}\]

Answer 4

\[\Large \lim_{h\to 0}\frac{f(1+h)}{h}=5\]
\[\Large \lim_{h\to 0}\frac{f(1+h){\LARGE \color{red}{-0}}}{h}=5\]
\[\Large \lim_{h\to 0}\frac{f(1+h){\LARGE \color{red}{-f(1)}}}{h}=5\]

so we see that \(\Large f(1) = 0\)

Answer 5

so my step would be invalid?

Answer 6

it's valid, I just don't see where to go with it

Answer 7

oh ok. And you knew to put in that f(1) as 0 simply because the equation resembled the derivative limit?

Answer 8

I saw it match up with the limit definition. Well almost match up. It was just missing the `-f(1)` part

Answer 9

so f(x) = 0 and f'(x) = 5?

Answer 10

f ' (1) = 5 and f(1) = 0

we can't say anything about f(x) or f ' (x) in general

Answer 11

ok, thank you

Answer 12

IF it was \[\Large \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=5\] then we can say f ' (x) = 5

Answer 13

ok got it