Question

The linear approximation at x = 0 to sqrt { 7 + 6 x } is A + B x where A is: ? and where B is: ?

Answer 1

Using, Taylor expansion, try please, we have:
\[A=\sqrt{7}, \] and \[B=\frac{ 3 }{ \sqrt{7} }\]

Answer 2

What are you studying for this? Have you heard of Taylor expansions yet?

Answer 3

I'm in calc 1. And the Taylor expansion, not sure what that is exactly.

Answer 4

I'm sorry, I don't know any other method other than Taylor expansion, in order to solve your exercise!

Answer 5

hmm. ok, well, what are you covering in class currently @cjroyg, so we can see what you are using if not taylor's

Answer 6

differentiation, linear approximation, and graphing rational functions

Answer 7

what are you using for linear approximations?

Answer 8

look familiar? http://tutorial.math.lamar.edu/Classes/CalcI/LinearApproximations.aspx

Answer 9

yes it does

Answer 10

alrighty then, so, first step, you want to find the derivative of \(\sqrt{7+6x}\) Do you know how?

Answer 11

yes, it would give you f'(x) = 3/sqrt(6 x+7)

Answer 12

then we will use this for our approximation \[L(x)=f(a)+f ~'~(a)(x-a)\]

Answer 13

your a is 0

Answer 14

plug and chug, and you are done

Answer 15

so a is my x value?

Answer 16

a is the point you are approximating around, which in this case is 0

Answer 17

the x in the eq is left alone when you plug values in, You only use the a value

Answer 18

so for the (x-a) it is actually (0-0), right?

Answer 19

no, leave the x there. We do not have an x value, just an a value

Answer 20

but it says at x=0 in the question?

Answer 21

well, that was what I was saying before, it wants the approximation around x=0, so technically, x never equals zero. Which is why we have a=0. We essentially have no x value

Answer 22

okay, I get it now.

Answer 23

ok, so, can you type in here what L(x)=?

Answer 24

So L(x) = (3 x)/sqrt(7)+sqrt(7)

Answer 25

how did you get that?

Answer 26

from L(x)= (sqrt(7+6(0)))+(3/sqrt(6 (0)+7))(x-0)

Answer 27

ah you flipped it, sorry

Answer 28

yep, I agree

Answer 29

now, cool thing. Notice that @Michele_Laino , has the same answer. This is because whatyou use for a linear approximation is actually the Taylor series evaluated at n=1

Answer 30

Nice.

Answer 31

yep yep, so do you understand?

Answer 32

btw, nice horse

Answer 33

I do understand, thanks for the help and for the horse. I love her so much, best pony I ever owned.

Answer 34

Nice! I haven't rode since I was thrown very badly few years ago. I miss it terribly though!

Answer 35

I understand. I now am at university so I only get to see her every couple months, and it is not easy. Especially that my last year at home, I also got into a bad fall from a green horse I was riding for a friend, so I did not get to ride her much before leaving.

Answer 36

:( that is no fun at all