Question

Linear approximation :D

Answer 1

\[e^{-0.015}\]

Answer 2

you want linear approximation of \(\large e^{-0.015}\), doesn't it involve x term?

Answer 3

no, we are just learning , I know the formula is f(a)-f'(a)(x-a)

Answer 4

Yeah, you're right

Answer 5

\[e^{-x}\to 0,\ if\ x\to \infty\]

Answer 6

if x approaches infinity, then e^(-x) approaches 0

Answer 7

use calculator to find this, just so you know
\[e^{-0.015}>>1\]
if it's not mentioned, then choose "a" anything, preferably a=0

Answer 8

yeah, what's your function?

Answer 9

What's f(x)?

Answer 10

Just a moment

Answer 11

ok :)

Answer 12

Ok, I got it. Let the function be
\[f(x)=e^{-x}\]
we have to find f(0.015)
Let's approximate it linearly
\[f(x)=f(a)+f'(a)(x-a)\]
a=0
\[f(x)=1-(x-0)\]
\[f(x)=1-x\]
do you get this?

Answer 13

\[e^{0}=1\]

Answer 14

umm

Answer 15

what function did you assume?

Answer 16

yeah, you should assume the function as e^(-x)
then linear approximation and then approximate the value

Answer 17

We could check also
\[e^{-0.015}=0.985\ \ \text{this is the exaxt value}\]
Now we'll check using our approximation
\[f(x)=1-x\]
\[f(0.015)=1-0.015=0.985\]
see we got same value :D

Answer 18

ahhhh i got it now, thanks @ash2326 !!!!

Answer 19

you're welcome :)

Answer 20

@ash2326 hey sorry to bug you again, but can you explain how i get f'(a)?

Answer 21

hello @amistre64 , ash was helping me earlier, and I was wonderring how he set up f '(a)

Answer 22

my function is f(x)=e^-x

Answer 23

at a = 0

Answer 24

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

Answer 25

\[e^0+ f ' a(x-0)\]

Answer 26

\[1+f'(a)(x)\]

Answer 27

how is f '(a) = -1?

Answer 28

why would you make x=0 in the first one, and not also in the second one?

Answer 29

a=0

Answer 30

@hartnn hey would you mind taking a quick look at this problem?

Answer 31

\[e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\]
the linear part ofthis is just\[e^x=1+x\]

Answer 32

regarding f'(x)
f(x)= e^{-x}
f'(x) = e^{-x} d/dx(-x) = -e^{-x}
when you put x=0
f'(0)=-1

Answer 33

@hartnn thank you!!!!!!!!!!

Answer 34

welcome ^_^

Answer 35

:)