Question

Find the linear approximation of f(x)=ln x at x=1 and use it to estimate ln 1.16.
L(x)= .
ln 1.16 approx

Answer 1

do you know how to find the tangent line to the curve f(x)=ln(x) at x=1?

Answer 2

no, i'm sorry

Answer 3

The tangent line to the curve y=f(x) at x=a is
\[y-f(a)=f'(a) \cdot (x-a)\]

Answer 4

does that look familiar?

Answer 5

If f(x)=ln(x) then f'(x)?

Answer 6

f'(x)=?

Answer 7

1/x?

Answer 8

good now what is f'(1)

Answer 9

if f'(x)=1/x then f'(1)=1/1 right?
so the tangent line is
\[y-f(1)=f'(1)(x-1) \\ y-\ln(1)=\frac{1}{1}(x-1)\]

Answer 10

I will let you finish simplifying that

Answer 11

\[\frac{1}{1}\]

Answer 12

\[y-f(a)=f'(a) \cdot (x-a) \\ y=f'(a)(x-a)+f(a) \\ f(x) \approx f'(a)(x-a)+f(a) \text{ for values near } x=a\]

Answer 13

you plug it in for a?

Answer 14

i'm sorry if I misunderstood

Answer 15

no \(a=1\) replace \(x\) by \(1.16\)

Answer 16

I still understand i'm so sorry

Answer 17

Dont*

Answer 18

what part is hard to understand

Answer 19

of finding ln1.16 approx

Answer 20

\[f(1.16) \approx f'(1)(1.16-1)+f(1)\]

Answer 21

we already found f'(1) and f(1)

Answer 22

fprime(1) is 1 and f(1) is ln(1) right?

Answer 23

yes but ln(1)=?

Answer 24

0 i believe

Answer 25

yes

Answer 26

so its 0.16?

Answer 27

so you have f(1.16) is approx 1.16-1

Answer 28

yes

Answer 29

then l(x)= the same thing?

Answer 30

what is l(x)?

Answer 31

oh is that 1*x
1*x is x

Answer 32

i don't know what you meant about it equaling the same thing though

Answer 33

oh i thought the answer was the same because it 2 separate questions

Answer 34

i don't understand what we are talking about anymore

Answer 35

lmao i'm sorry, if you look back to the question, they're asking for 2 separate answers

Answer 36

their asking what is L(x) and what is ln1.16

Answer 37

We already found the tangent line to x=a
and then used it to approximate ln(1.16)

Answer 38

tangent line to f at x=a*

Answer 39

our a was 1

Answer 40

I'm sorry, so what is l(x) you probably stated this in your question

Answer 41

\[y-f(a)=f'(a)(x-a) \\ y=f'(a)(x-a)+f(a) \\ L(x)=f'(a)(x-a)+f(a) \\ f(x) \approx L(x) \\ f(x) \approx f'(a)(x-a)+f(a)\]

Answer 42

L(x) is f'(a)(x-a)+f(a)

Answer 43

but this was already found above

Answer 44

remember f'(x)=1/x
f'(1)=1
f(1)=ln(1)=0

Answer 45

and what is a

Answer 46

nvm i know a

Answer 47

\[L(x)=(\frac{1}{x})_{x=1}(x-1)+\ln(1)\]

Answer 48

(1/x) evaluated at x=1 is of course 1/1=?

Answer 49

its x-1

Answer 50

yes

Answer 51

Omg

Answer 52

lol

Answer 53

I'm so sorry i feel like i frustrated you so much

Answer 54

like that is what we plugged 1.16 into

Answer 55

Thank you for all your help. Again i'm sorry

Answer 56

to approximate ln(1.16)

Answer 57

np

Answer 58

I appreciate it a lot, i know it's frustrating to teach me

Answer 59

nah not really

Answer 60

Thanks :)