# Linear Approximation

1. Mar 7, 2017

### mastermechanic

<Moved from a technical section and thus a template variation>

1-) Question:
Let f, g and h be differentiable everywhere functions with h(1) = 2 , h'(1) = - 3 , g(2) = -1 , g'(2) = 5 , f(-1) = 4 , f'(-1) = 7. Approximate the value of function F(x) = f(g(h(x))) at point x= 1.001

2-) My attempt: I think it's linear approximation question. I've generated the main functions by assuming their derivative as their slope.

If $$h(1)=2 , h'(1) = -3 , (1,2) h(x) = -3x+5$$
If $$g(2) = -1 , g'(2) = 5 , (2,-1) g(x) = 5x-11$$
If $$f(-1) = 4 , f'(-1) = 7 , (-1,4) f(x) = 7x+11$$

So $$F(x) = f(g(h(x))) = [ 7[ 5 (-3x + 5) - 11 ] +11]$$

$$F(x) = -105x + 109$$
$$F(1.001) = 3.895$$

Is it correct?

Thanks.

Last edited: Mar 7, 2017
2. Mar 7, 2017

### phyzguy

I think your method is correct, but you've made at least two mistakes. Try checking your expression for F(x) by calculating F(1) and comparing it to f(g(h(1))), which you know. Also, how did you get from your expression for F(x) to the final answer?

3. Mar 7, 2017

### FactChecker

It looks like the question is set up for an application of the chain rule. Are you supposed to know the chain rule at this point or is this question preparation for introducing it.

4. Mar 7, 2017

### mastermechanic

I reached the final equation by putting each function to inside other one.
Thanks for your caution, I've realized that I made some multiplication mistake. When I fixed it , I got $$F(x) = -105x + 109$$ and if we put the x=1.001 value get 3.895 which totally make sense now, because as you said when I assumed x as 1 h(1) = 2 g(2) = -1 and f(-1) = 4 so it is too close.

Thanks :)