anonymous
  • anonymous
Calculus: Evaluate f(x)=1/x at the specified value of the independent variable and simplify the result. (f(1+Δx)-f(1))/Δx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[f(x)=1/x\] \[\frac{ f(1+Δx)-f(1) }{ Δx }\]
anonymous
  • anonymous
@Loser66
Loser66
  • Loser66
@DebbieG

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anonymous
  • anonymous
Um... does anyone know to how to solve this?
DebbieG
  • DebbieG
So is this a differential?
anonymous
  • anonymous
I just started my calculus class, so we're reviewing with substitutions
DebbieG
  • DebbieG
Dont you just treat the Δx like we do the "h" in the difference quotient? (I'm not terribly familiar with differentials... but isn't it just a riff on the difference quotient?)
DebbieG
  • DebbieG
ok.. well... my best guess is to treat is like a diff quotient. Just use that Δx in place of h. So you need to use the function: \(\Large \dfrac{ f(1+Δx)-f(1) }{ Δx }=\dfrac{\dfrac{1}{1+Δx}-1 }{ Δx }\) and then simplify that....
DebbieG
  • DebbieG
unless you know what Δx is...if you were given another x value somewhere?
anonymous
  • anonymous
no the problem is as I wrote it
DebbieG
  • DebbieG
Differentials are weird. I guess it would go something like this?.... (but I'm not terribly confident): \(\Large \left(\dfrac{1}{1+Δx}-1 \right)\cdot\dfrac{1}{ Δx }\\ \Large \left(\dfrac{1-(1+Δx)}{1+Δx} \right)\cdot\dfrac{1}{ Δx }\\ \Large \left(\dfrac{-Δx}{1+Δx} \right)\cdot\dfrac{1}{ Δx }\\ \Large \dfrac{-1}{1+Δx} \)
DebbieG
  • DebbieG
but it kind of makes sense, since as Δx->0 that expression goes to -1, and the derivative here \(f\prime(1)=-1\).
anonymous
  • anonymous
I don't understand how you went from step 1 to step 2
DebbieG
  • DebbieG
@southpaw, I'm sorry, I had to run out unexpectedly.... All I did from step 1 to step 2 was to join everything inside the () over the common den'r.... \(\Large \Large \left(\dfrac{1}{1+Δx}-1 \right)\cdot\dfrac{1}{ Δx }\\ \Large \left(\dfrac{1}{1+Δx}-\dfrac{1+Δx}{1+Δx} \right)\cdot\dfrac{1}{ Δx }\\ \Large \left(\dfrac{1-(1+Δx)}{1+Δx} \right)\cdot\dfrac{1}{ Δx }\\ \Large \left(\dfrac{-Δx}{1+Δx} \right)\cdot\dfrac{1}{ Δx }\\ \Large \dfrac{-1}{1+Δx}\)
anonymous
  • anonymous
alright thank you

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