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## anonymous one year ago ???

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1. freckles

you can find the tangent line at x=e the only problem is when differentiating f(x)=2ln(x) and f'(e) is probably something you would want to use calculator for but I was told the whole point of these exercises was to approximate without using calculator I guess we could just use the approximate for e as 2.72 when it comes to that anyways the tangent line at x=e will look something like this: y-f(e)=f'(e)(x-e) y=f(e)+f'(e)(x-e) so f(x) is approximately f(e)+f'(e)(x-e)

2. anonymous

So I would plug in values into that?

3. freckles

you can use that approximation equation to approximate f(2) f(2) is approximately f(e)+f'(e)(2-e)

4. freckles

you still need to find f(e) and f'(e) though

5. anonymous

How would I find those. And would it help if i showed the answer choices?

6. freckles

there are choices for approximating f(2)?

7. freckles

f(e) can bound by pluggin in e into f(x)=2ln(x) f'(x) needs to be found before finding f'(e)

8. anonymous

y = x − 2 y = x − 2 + ln4 y = 2ln(x − 2) y = 1 This is what it shows

9. freckles

oh they want the linear approximation for f(2)

10. freckles

they want the tangent line at x=2

11. freckles

so just find the tangent line at x=2

12. anonymous

When I derive 2lnx I get 2/x

13. freckles

right now plug in 2

14. freckles

$y-f(2)=f'(2)(x-2) \\ y=f(2)+f'(2)(x-2)$

15. anonymous

I got 1 when I plugged it in

16. freckles

ok so f'(2)=1 you still need to find f(2)

17. freckles

$y=f(2)+1(x-2) \\ y=f(2)+(x-2) \\ y=f(2)+x-2 \\ y=x-2+f(2)$

18. freckles

f(x)=2ln(x) input 2 to find f(2)

19. anonymous

1.39

20. freckles

don't approximate plug in 2

21. freckles

then just use power rule for log

22. freckles

$a \ln(x)=\ln(x^a) \text{ power rule }$

23. freckles

$f(2)=2\ln(2)=...$

24. anonymous

When I plugged that into my calculator I got 1.39

25. freckles

yes that is great but we are not approximating use power rule bring the 2 up as an exponent of what is inside the log

26. freckles

example $5\ln(2)=\ln(2^5)=\ln(32)$

27. freckles

you try 2ln(2)=what using power rule

28. anonymous

So I would do ln(2^2)=ln(4)

29. freckles

yes

30. freckles

the tangent line at x=a looks like this y-f(a)=f'(a)(x-a) so we has 2 instead of a y-f(2)=f'(2)(x-2) you found f(2) as ln(4) and f'(2) as 1 y-ln(4)=1(x-2) y-ln(4)=x-2 add ln(4) on both sides and you should see clearly see your answer

31. anonymous

x-2+ln4?

32. freckles

yes

33. freckles

y=x-2+ln(4)

34. anonymous

Thank you so much!!!!

35. freckles

np

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