Description
i need help with all parts of question 3 and 4. here is a pic of the problems
please include all work in the answers. thanks in advance
Explanation & Answer
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
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