Rizags
  • Rizags
If f is differentiable at 1, and the limit of f(1+h)/h as h->0 is 5, what is f(1)?
Mathematics
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SOLVED
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chestercat
  • chestercat
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Rizags
  • Rizags
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
` 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
Rizags
  • Rizags
so \[f'(1) = 5-\lim_{h \rightarrow 0}\frac{f(1)}{h}\]

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jim_thompson5910
  • jim_thompson5910
\[\Large \lim_{h\to 0}\frac{f(1+h)}{h}=5\] \[\Large \lim_{h\to 0}\frac{f(1+h){\LARGE \color{red}{-0}}}{h}=5\] \[\Large \lim_{h\to 0}\frac{f(1+h){\LARGE \color{red}{-f(1)}}}{h}=5\] so we see that \(\Large f(1) = 0\)
Rizags
  • Rizags
so my step would be invalid?
jim_thompson5910
  • jim_thompson5910
it's valid, I just don't see where to go with it
Rizags
  • Rizags
oh ok. And you knew to put in that f(1) as 0 simply because the equation resembled the derivative limit?
jim_thompson5910
  • jim_thompson5910
I saw it match up with the limit definition. Well almost match up. It was just missing the `-f(1)` part
Rizags
  • Rizags
so f(x) = 0 and f'(x) = 5?
jim_thompson5910
  • jim_thompson5910
f ' (1) = 5 and f(1) = 0 we can't say anything about f(x) or f ' (x) in general
Rizags
  • Rizags
ok, thank you
jim_thompson5910
  • jim_thompson5910
IF it was \[\Large \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=5\] then we can say f ' (x) = 5
Rizags
  • Rizags
ok got it

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