Question

The linearization at a = 0 to sqrt(5 + 9x) is A + Bx where A is ____ and B is ____?

Answer 1

The binomial expansion\[(1+x)^\alpha=1+\alpha x+\frac{a(a-1)}{2!}x^2+\mathrm O(x^3)\]
For a linear approximation\[(1+x)^\alpha=1+\alpha x+\mathrm O(x^2)\]

Answer 2

this looks greek

Answer 3

yep,

can you rearrange

sqrt(5 + 9x) to be of the form (1+y)^a

Answer 4

I do not understand how, what is the "O"?

Answer 5

the O is the order of the terms that we are ignoring,

With the linearisation at the point x=0, O(x^2) tends to 0 so we can ignore them

maybe this expression is more readable
\[(1+x)^\alpha\approx1+\alpha x\]

Answer 6

okay let me try

Answer 7

what would I put into x for in "(1+x)^a?

Answer 8

could I use this? y - f(a) = f '(a) (x-a)

Answer 9

\[\sqrt{5 + 9x} = (5 + 9x)^{1/2}=\left(5(1+\tfrac95x)\right)^{1/2}=\sqrt5(1+\tfrac95x)^{1/2}\]

Answer 10

Okay, so got that, how do I find A and B?

Answer 11

\[(1+\tfrac95x)^{1/2} \to (1+y)^{\alpha} \approx1+\alpha y\]

Answer 12

\[\sqrt5(1+\tfrac95x)^{1/2} \to \sqrt5(1+y)^{\alpha} \approx\sqrt5(1+\alpha y)\]

Answer 13

is that "a" or something else? alpha?

Answer 14

I am confused

Answer 15

@johnnydicamillo, it seams like the method i'm trying to explain, is different to the method you have learnt. If you close this question and ask it again, i think someone who knows the right method will find this question .

Sorry for the confusion .