What is the Correct Linearization of a Function?

[Qube]

Homework Statement

http://i3.minus.com/jbt2vueBfwXvWD.jpg

Homework Equations

Linearization: f(x) + f'(x)(dx)

The Attempt at a Solution

The derivative of g(x) using the chain rule is (2lnx)/x. x = e, so that simplifies to 2/e.

Linearization:

(2/e)x + 1, where 1 is f(e).

Why is the answer (2/e)x - 1 instead? I thought f(e) is 1, not negative 1?

 

[tiny-tim]

Hi Qube!

That works fine for x = 0 … not so much for x = e !

(2/e)x + 1 at x = e is 3

(2/e)x - 1 at x = e is 1 …

which do you prefer? ​

(2/e)(x - e) + 1 would be ok

 

[eumyang]

Linearization: f(x) + f'(x)(dx)

I was told that the linearization is defined this way:
$L(x) = f(a) + f'(a)(x - a)$,
where f is differentiable at a.

 

[Qube]

Hi Qube!

That works fine for x = 0 … not so much for x = e !

(2/e)x + 1 at x = e is 3

(2/e)x - 1 at x = e is 1 …

which do you prefer? ​

(2/e)(x - e) + 1 would be ok

Yes, I see how working backwards helps, but how can I work to the solution instead of vice versa?

 

[Qube]

I was told that the linearization is defined this way:
$L(x) = f(a) + f'(a)(x - a)$,
where f is differentiable at a.

Yes, and I seem to get 1 + (2/e)x

 

[tiny-tim]

$L(x) = f(a) + f'(a)(x - a)$

f(a) = 1, f'(a) = 2/e, (x - a) = (x - e)

total: 1 + 2/e (x - e) = (2/e)x - 1

 

[Mark44]

Show us what you did.

L(x) = g(x) + g'(a)(x - a)

For L(e), what do you get? I get (2/e)x - 1.

 

[Qube]

Ohh I see. Alright.

a = e

f(e) = 1

f'(e) = 2/e

1 + (2/e)(x-e) = 1 + (2/e)x - 2 and the 1 - 2 results in the -1. Alright. I think my problem was not sticking closely enough with the definition and instead going with a gut feel / intuition about how linearization works.

 

[tiny-tim]

… and instead going with a gut feel / intuition about how linearization works.

the gut is not linear! …