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I'm a little confused by this question:

Lines and things that are linear are relatively boring in mathematics. What if my function f(x) = g'(x). I’m going to ask the same question in a different way.

  1. What is an approximate value of g'(1.1)?

So I know $L(x) = f(a) + f'(a)(x-a)$

If I substitute items in, I get $g'(x) + g''(x)(x-x)$.

Since $x-x = 0$ I always am going to end up with $L(x) = g'(x)$

I don't think this can be the right answer since I know the whole point is to use the 2nd derivative, but if you get $0$ for $(x-a)$, then the whole part ends up being $0$. What am I missing here?

EDIT: I'm completely lost here because these are the followup question after the one above:

Now things are interesting. Suppose g(1) = 0 and g'(1) = 1.

  1. Use linear approximation to approximate the value of g(1.1). The answer to #2 and #3 gives us a slope of a tangent line and a value of a function, which we can use.

  2. Use your previous answers to approximate the value of g(1.2). Somehow, this second derivative has caused an adjustment to our approximation.

  3. What did using the second derivative do to the approximation?

  4. Do you believe the approximation is better or worse than our original? Why?

Also...MathJax isn't working fo me...it still is showing the $ in here

Lorenzo B.
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MattE
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  • Welcome to MSE. Please edit and use MathJax to properly format math expressions. – Lee David Chung Lin Mar 10 '19 at 16:13
  • $g'(1.1)\approx g'(1)+g''(1)\times(1.1-1)=f(1)+f'(1)\times 0.1$ – J. W. Tanner Mar 10 '19 at 16:16
  • $g(1.1)\approx g(1)+g'(1)\times(1.1-1)=g(1)+f(1)\times0.1$ – J. W. Tanner Mar 10 '19 at 18:56
  • $g(1.1)\approx g(1)+g'(1.1)\times (1.1-1)\approx g(1)+f(1)\times0.1+f'(1)\times0.1^2$ – J. W. Tanner Mar 10 '19 at 19:00
  • @J.W.Tanner We are suposed to find values for these, not just give equations – MattE Mar 10 '19 at 19:16
  • I wasn't going to do all your work; in your edit you gave values for $g(1)$ and $g'(1)$, which could easily be plugged in to the equations I gave – J. W. Tanner Mar 10 '19 at 19:19
  • @J.W.Tanner Ahh! OK got it...I guess I am just tryign to understand the whole second derivative part of what he is saying...just a little confusing to me. Admittedly Calculus is not my most favorite subject – MattE Mar 10 '19 at 20:01
  • Why did you reproduce 2., 3., 4., 5., and 6., but not 1.? – J. W. Tanner Mar 10 '19 at 20:07
  • Because 1(and 2) was an easy question that only required the use of the linear approximation formula, which I completed myself. Question 2 here was a "different way" of asking the linear approximation of f(1.1) and the subsequent questions relied on it...just was a little confused by using a second derivative. Also...where does the 0.1 squared come in in the last formula? – MattE Mar 10 '19 at 20:13
  • to get the third formula, I substituted the first formula for $g'(1.1)$, and the result had $0.1^2$ – J. W. Tanner Mar 10 '19 at 20:47
  • @J.W.Tanner The way he forms the question, I am thinking he wants us to somehow use the slope and the equation in part 4..so I know the slope is the first derivative, which would be g'(1), which equals 1. And the equation for g(1.1) ends up being g(1.1) = x-1. I know the equation for a slope is y = mx+b. I believe if I am plugging in values I get 1 = 1(1.2) +b and that makes b = -0.2. But I don't know exactly how to use this to find 1.2 from 1.1's slope and formula other than just using the linear approximation of 0.2 which I am pretty sure is not what he is looking for – MattE Mar 10 '19 at 20:55

1 Answers1

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Given $g(1)=0$ and $g'(1)=1$,

$g(1.1)\approx g(1)+g'(1)(1.1-1)=0+1\times 0.1=0.1$ and

$\color{blue}{g'(1.1)\approx g'(1)+g''(1)\times (1.1-1)=1+0.1g''(1).}$

Using linear approximation, $g(1.2)\approx g(1)+g'(1)(1.2-1)=0+1\times 0.2=0.2.$

Better approximation:

$g(1.2)\approx g(1.1)+\color{blue}{g'(1.1)}(1.2-1.1)\approx 0.1 + \color{blue}{\left(1+0.1g''(1)\right)}0.1 = 0.2 + 0.01g''(1)$

J. W. Tanner
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