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##### Calculous rate of change problem 2

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i need help with all parts of question 3 and 4. here is a pic of the problems

http://imgur.com/F44GY9Z

Jan 19th, 2015

f(x)=100,000+2000t2

f’(x)= 4000t

f’(10) = 4000*10 =40000= 40,000

 x h f(x) Diff quot 10 0.1 304020 10 300000 40200 10 0.5 320500 10 300000 41000 10 0.05 302005 10 300000 40100 10 0.01 300400.2 10 300000 40020 10 0.001 300040 10 300000 40002 10 0.0001 300004 10 300000 40000.2

We can see the diff quotient approaches 40000

g(x) =1/x

Tangent x= 1 and x= -2

Slope of tangent at x=1 is g’(x) at x=1

g’(x) = -1/x2 ;  g’(1) = -1

g’(-2) = -1/(-2)2  =-1/4

Equation of the tangent at x=1  and y =1/x =1is

(y-1)= -1(x-1)

y-1+x-1=0

y+x =2

or y =-x+2

Equation of the tangent at x= -2 and y =-1/2 is

(y-(-1/2)) = -1/4(x-2)

y+1/2 = -1/4 *x+1/2

y = -x/4

[ f(1+h)-f(1)]/h   = [1/(1+h) -1] /h

=(1-(h+1))/[h(h+1)]= -h/[h(h+1)] =-1/(h+1) -> -1 as h->0

[ f(-2+h)-f(-2)]/h   = [1/(-2+h) -1/(-2)] /h

= [1/(-2+h) +1/2] /h

=(2+1(-2+h))/[2(-2+h]= h/[h[2(-2+h)] =

-1/[2(-2+h)] -> -1/4  as h->0

Jan 19th, 2015

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Jan 19th, 2015
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Jan 19th, 2015
Sep 20th, 2017
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