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

  • $\begingroup$ Welcome to MSE. Please edit and use MathJax to properly format math expressions. $\endgroup$ – Lee David Chung Lin Mar 10 at 16:13
  • $\begingroup$ $g'(1.1)\approx g'(1)+g''(1)\times(1.1-1)=f(1)+f'(1)\times 0.1$ $\endgroup$ – J. W. Tanner Mar 10 at 16:16
  • $\begingroup$ $g(1.1)\approx g(1)+g'(1)\times(1.1-1)=g(1)+f(1)\times0.1$ $\endgroup$ – J. W. Tanner Mar 10 at 18:56
  • $\begingroup$ $g(1.1)\approx g(1)+g'(1.1)\times (1.1-1)\approx g(1)+f(1)\times0.1+f'(1)\times0.1^2$ $\endgroup$ – J. W. Tanner Mar 10 at 19:00
  • $\begingroup$ @J.W.Tanner We are suposed to find values for these, not just give equations $\endgroup$ – MattE Mar 10 at 19:16

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:

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


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