# Linearization of a function

1. Oct 28, 2013

### Qube

1. The problem statement, all variables and given/known data

http://i3.minus.com/jbt2vueBfwXvWD.jpg [Broken]

2. Relevant equations

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

3. 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?

Last edited by a moderator: May 6, 2017
2. Oct 28, 2013

### 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

3. Oct 28, 2013

### eumyang

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

4. Oct 28, 2013

### Qube

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

5. Oct 28, 2013

### Qube

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

6. Oct 28, 2013

### tiny-tim

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

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

7. Oct 28, 2013

### Staff: Mentor

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.

8. Oct 28, 2013

### 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.

9. Oct 28, 2013

### tiny-tim

the gut is not linear!