Question

How to approximate a number using f(x)=-1/4x+3/2

Answer 1

It wants the approx of \[\sqrt{3.9}\] and i know i need to plug it in for x but i dont know how

Answer 2

It wants the approximate of it, so i cant just plug in \[\sqrt{3.9}\]. i should get the decimal approx of it using this formula.

Answer 3

A linear approximation... like this?\[
f(x_2)=f'(x_1)(x_2-x_1)+x_1
\]

Answer 4

yes.

Answer 5

Let \(x_1=4\) since it's easy to find the square root of it, and it's really close to \(3.9\)

Answer 6

Should have written: \[

f(x_2)\approx f'(x_1)(x_2-x_1)+f(x_1)

\]

Answer 7

Can i plug it in for x using the simplified form?

Answer 8

no

Answer 9

First of all, what is \(f'(x)\) in this case?

Answer 10

is there any reason to put 2-(1/4)(x+2) into f(x)=-1/4x+3/2

Answer 11

I have no idea wtf you're doing. You come outta nowhere with these equations.

Answer 12

f'(a)= -(1/4)
f(a)= 2

Answer 13

sorry lol. Im doing an linear approx of f(x)= radical(2-x) at a=-2

Answer 14

Okay, you just need to do \(x_2-x_1\)...
What is the problem?

Answer 15

I have it written as f(x)=f(a)+f'(a)(x-a)

Answer 16

So you're using a linear approximation at \(4\).

That is a statement. What is the question?

Answer 17

I dont understand what you are saying.

Answer 18

I really am trying. its just not making sense. you are plugging in 4 for f(a)?

Answer 19

no that cant be right. it should be plugged in for x

Answer 20

I just cant get the right approx when i plug 4 in for x.

Answer 21

maybe i did the values wrong.

Answer 22

\[
f(x_2)\approx f'(x_1)(x_2-x_1)+f(x_1)
\]We have \(x_1=4,x_2=3.9\) \[
f(3.9)\approx f'(4)(3.9-4)+f(4)
\]

Answer 23

Can you do that?

Answer 24

\[f(3.9)=3.6\]

Answer 25

its i got it. you just take 4-3.9= 0.1
then you plug it in.