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May 20th, 2015, 10:30 AM   #1
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Linear Approximation Clarification

Hey Guys,

So I have 2 homework questions I didn't really understand. They are question 34 C and 35 B.

I attached the question and solution to the two questions below, but I didn't really understand the solution. Here: Gyazo - faf58e848ac1eef80ed3a0488741a11c.png


What I understood is:

34.C How I understood it is that the vertex of 1-x^2 at x=0 is lying directly on top of the function 1/(1+x^2). In essence, both functions intersect at (0,1). Hence, the slope of the tangent of the vertex 1-x^2 is the same as the slope of the tangent at 1/(1+x^2) (ie, 0). Is that true?



35.B
With local linearization, we've been creating lines tangent to a point on a function in which we use to estimate values around to it. The answer in A is linear, yet B is a quadratic, meaning the local linearization we've been looking at isn't so linear anymore. Is the idea behind this that this quadratic is a much better approximation of points close to x=0 (than a line)?

Also, why is it that if g(x)=f(x^2) than we can get a quadratic g(x) that can approximate f(x) as 1-2(x^2) just by subbing in (x^2)?

I did it by hand and got a very different answer with very different approximations

l(x)=g'(x)(x-a)+g(x)

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

g'(0)=0

g(0)=1

l(x)=0*(x-0)+1 --> l(x)=1 which is very different than the answer 1-2(x^2) if -0.25>x or x>0.25.

(But a good approximation for values very close around x=0) (Gyazo - 75becba1945f9fe397b8e33bd4df4d05.png)

Is the idea behind this question that the real linear approximation is just a horizontal line y=1 but if you didn't want to do all the work 1-x^2 (which is g(x)=f(x^2)) would be a reasonable estimate?

From,
Tzad

Last edited by skipjack; May 20th, 2015 at 01:57 PM.
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