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The limit below represents a derivative \(f'(a)\). Find \(f(x)\) and \(a\).

 

\(\lim_{h\to 0}\frac{\left(2+h\right)^{2}-4}{h}\)

 

\(f(x) =\) 

 

\(a = \)

 Feb 24, 2022
 #1
avatar+122 
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Idk how to do this

 Feb 24, 2022
 #2
avatar+117224 
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Here is a pic for differentiation from forst principals.

 

The differential of a curve at a particular point is the gradient of the curve at that point.

A secant is a line that joins two poins on the curve.  In this case the orange line is a secant.

the gradient of the secant is   \(\frac{rise}{run}=\frac{change \;in \;y\; values}{change\; in\; x\; vlaues}=\frac{f(a+h)-f(a)}{(a+h)-a}=\frac{f(a+h)-f(a)}{h} \)

 

If you make h smaller and smaller then you are letting h tend to 0 and once h is 0 you will have the gradient of the tangent at that x point.    

\(gradient \;of \;tangent \;at\;x=a\\ y'(a)= \displaystyle \lim_{h\rightarrow 0} \frac{f(a+h)-f(a)}{h}\)

 

Most people just memorize this formula without necessarily understanding where it came from.

But for the smarter students it will help a lot of you understand it.

 

 

Lets compare this to what you have in your question

\(y' (a)= \displaystyle \lim_{h\rightarrow 0} \frac{ f(a+h)-f(a)}{h}\\ y' (a)= \displaystyle \lim_{h\rightarrow 0} \frac{(2+h)^2 -4}{h}\\ \)

 

so

  \(f(a)=4\\ f(a+h)=(2+h)^2   \;\;so\\ a=2 \;and\;\\ f(whatever)=(whatever)^2 \;\;so \\ f(x)=x^2 \;\;f(2)=2^2=4 \;\;(\text{which I already knew})\\~\\\)

 Feb 25, 2022
edited by Melody  Feb 25, 2022

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