linear approx.

[Guest]

Hi...
The problem is to find the linear approx of y=f(x)=1/(2x) @ a = 3...

Here's what I have so far:

f'(x) = -1/(2x^2)
f'(3) = -1/(2(3)^2 = 1/(2*9) = 1/18

so the linear approx is:

L(x) = f(a)+f'(a)(x-a)
L(x) = f(3) + f'(3)(x-3)
= (1/6) + (x-3)/3
= -5/6 + 1/3x??????????

Does that look correct? Thanks in advance its my first one of these!

 

[tkhunny]

This problem statement makes no sense. One can EASILY calculate the exact value of f(3). Why, then, would ANY approximation be necessary?

The usual idea is to approximate a value that we cannot calculate easily, but is NEAR a value that we can calculate easily. For example, sqrt(24) is a little tricky, but sqrt(25) is no problem. This probably explains why you have an 'x' left over.

 

[Guest]

Tk,
Thank you for your reply. I understand what you are saying (I think ) about approximation... i.e. why would we want to approx something that we can solve definatively...
However, I doublechecked the problem and that is exactly how it is written. So am I doing the correct thing with the problem as its stated - at least the steps anyway? Thanks

 

[pka]

You may want an approximation: f(x+Δx)~f(x)+(Δx)f’(x).
f(3+Δx)~f(3)+(Δx)f’(3).
Thus if you want f(2.98)~f(3)+(−0.02)f’(3).