Please help...

G

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The limit lim ((2+h)^3-8) / h represents the deriv of some f(x) @ some
h->0 point a. Find f(x) and a.


How do you find the original function from the derivative?
 
lim f'[x-->2] =[ [2+h]^3-8]/h expand[ 2+h]^3
=[[2^3+3[2^2]h +3[2][2]h^2/2! +h^3 ]-8] /h
=[8 +12h+6h^2 +h^3-8] /h
=h[12-6h+h^2] /h lim as h-->0
=12

we know that thederivative at x=2 is 12
f [x+h]= [x+h]^3 because the x was replaced by 2
then f[x]= x^3

proof
f[x]=x^3
f'[x]=3x^2 at x=2
f'[2]=12 check
Arthur
 
Thanks Arthur for the reply...

Can you help me understand a little better...

Did you just randomly pick as x-->2? Why would you use 2 instead of 0 like the problem stated. Sorry for the questions, im just completely in the dark here!

Chris
 
the problem stated h =0 not x=0

The normal way to derive the derivative of a function of x is:

y= f[x]
y+dy =f[x+h] subtract first from second
y+dy -y = f[x+h] - f[x]
dy = f[x+h]-f[x] divide by h
dy/h = [ f[x+h]-f[x]] / h lim h-->0
y ' =lim [f[x+h] - f[x] ] /h

let us let f[x]=x^2 and find its derivative
f[x+h]=[x+h]^2 subtract first from second divide by h
f[x+h]-f[x] =[ [x+h]^2 -x^2] / h or in the limit as h-->0
f ' [x] =[ [x+h]^2-x^2]

this will come out to be f ' [x] = 2x as we did befor but let us find f ' [x] at x=2

f'[x] =[ [2+h]^2 - 2^2]] / h lim as h-->0 doesn't this look familiar?
the instructor gave you the derivation of the derivative at a point x=2.
you always determine the derivative by taking the value at x+ delta x
{which is reolaced by h} subtracting the valus at x and making the delta infinetly small or let it approach 0

hope this helps
Arthur
 
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