Consider the following function. RX)=ln(1 + 2x), a=3, n= 3, 2.7S x 3.3 (a) Appro

Question

Consider the following function. RX)=ln(1 + 2x), a=3, n= 3, 2.7S x 3.3 (a) Approximate fby a Taylor polynomial with degree n at the number a 2, 16 98 2058 (b) Use Taylor's Inequality to estimate the accuracy of the approximation x)T(x) when x lles in the glven Interval. (Round your answer to slx declmal places.) (c) Check your result in part (b) by graphing R(x)I 0.00002 28 2.9 3.0 3.1 2 3.3 -5. x 10-6 -0.00001 -0.000015 0.000015 0.00001 5. x 10-6 -0.00002 2.8 2.9 3.0 3.1 3.2 3.3 0.00002 28 2.9 3.1 3.2 3.3 0.000015 -5. x106

Explanation / Answer

f = ln(1 + 2x)

f(3) = ln(1 + 6) = ln(7)

Deriving :
f' = 2/(1 + 2x) = 2/(1 + 2*3) = 2/7

f'' = -4/(1 + 2x)^2 = -4/49

f''' = 16/(1 + 2x)^3 = 16/343

So, we have
f(a) + f'(a)*(x - a)/1! + ....

= ln(7) + 2/7*(x - 3) / 1! + (-4/49)(x - 3)^2/2! + (8/343)(x-3)^3/3!

= ln(7) + 2(x-3)/7 - 2/49(x-3)^2 + 8/1029(x-3)^3 -----> ANS

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b)
Rn(x) = M/(n + 1)! * |x - a|^(n + 1)
with |x - a| <= d

We are given 2.7 <= x <= 3.3

3 - 0.3 <= x <= 3 + 0.3

So, |x - 3| <= 0.3

And thus a = 3 and d = 0.3

Here n = 3 :
So,
R3(x) = M/4! * (x - a)^4

Now, M here is the max value of f4(x)
Clearly f4(x) = -96/(1 + 2x)^4
To max this, we gotta use min(x), i.e 2.7

So, M = 96/(1 + 2*2.7)^4
So, M = 0.057220458984375

And with this :
R3(x) = M/4! * (x - a)^4

R3(x) = 0.057220458984375/4! * (x - a)^4

= 0.057220458984375/4! * (2.7 - 3)^4

= 0.0000193119049072265625

So, R3(x) <= 0.000019 ----> ANS

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c)
We have f(x) = ln(1 + 2x)
And T3(x) = ln(7) + 2(x-3)/7 - 2/49(x-3)^2 + 8/1029(x-3)^3

So, |R3(x)| =
|ln(1 + 2x) - (ln(7) + 2(x-3)/7 - 2/49(x-3)^2 + 8/1029(x-3)^3)|

which comes out to be graph BOTTOM LEFT ----> ANS

Didnt work?
Try graph TOP LEFT

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