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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

please include all work in the answers. thanks in advance

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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