## anonymous one year ago If the local linear approximation of f(x) = 2cos x + e2x at x = 2 is used to find the approximation for f(2.1), then the % error of this approximation is greater than 15% between 11% and 15% between 5% and 10% between 0% and 4

1. anonymous

@jim_thompson5910

2. jim_thompson5910

how far did you get here?

3. anonymous

well i found the linear approximation for f(2.1) using x = 2

4. jim_thompson5910

what did you get for f ' (x)

5. anonymous

tell me if i did thi right f(2.1) = 53.76 + 107.38(x - 2) f(2.1) = 107.38x - 161

6. jim_thompson5910

I'm getting y = 107.37771x - 160.98955 for the tangent line

7. jim_thompson5910

let g(x) = 107.37771x - 160.98955 compute g(2.1) and compare that with f(2.1)

8. anonymous

g(2.1) = 64.503641 f(2.1) = 65.67663883

9. jim_thompson5910

now compute the percent error using those f(2.1) and g(2.1) values

10. anonymous

so just divide them by each other?

11. jim_thompson5910

$\Large \text{Percent Error} = \frac{|\text{Approximate Value}-\text{Exact Value}|}{\text{Exact Value}}$ $\Large \text{Percent Error} = \frac{|g(2.1)-f(2.1)|}{f(2.1)}$ $\Large \text{Percent Error} = ??$

12. anonymous

1.81

13. anonymous

are you sure it's not backwards on the numerator

14. anonymous

15. jim_thompson5910

the order in the numerator does not matter because |x - y| = |y - x|

16. jim_thompson5910

if the absolute value bars weren't there, then yeah, the order matters

17. anonymous

okay well thank you very much :D!!!

18. jim_thompson5910

I'm getting roughly 0.01786 which is approximately 1.786% error

19. anonymous

yeah me too either way its D

20. jim_thompson5910

yes