Question

The tangent line approximation for f(x)=sqrt (x^2+16) near x=-3 is...?

Answer 1

@amistre64 @ganeshie8

Answer 2

@zepdrix @hartnn

Answer 3

so what are the steps to do a tangent line approx ?
you know ?

Answer 4

it is the same as local linearization i think?

Answer 5

yes
\(f(x) \approx f(a) + f'(a) (x-a)\)

Answer 6

a =-3 here

Answer 7

ok. so a=-3, so f(x)=f(-3)+f'(-3)(x+3)

Answer 8

and \[f'(x)=\frac{ 2x }{ 2\sqrt{x+16} }\]

Answer 9

yes

Answer 10

if you mean x^2 +16

Answer 11

\[f(x)= \sqrt{7}-\frac{ 3\sqrt{7} }{ 8}(x+3)\]

Answer 12

sqrt 7 ?
8?
howw ?

Answer 13

i think my math's wrong, gimme a sec

Answer 14

ohh
(-3)^2 is not -9
its +9

Answer 15

oh uhggggggg I did -3^2 is -9

Answer 16

try again, you will get it this time :)

Answer 17

\[f(x)=5-\frac{ 3 }{ \sqrt{13} }(x+3)\]

Answer 18

I think that's right

Answer 19

13 :O ??

Answer 20

from where did 13 come ?

Answer 21

must I rationalize denominator?

Answer 22

oh my derivative is wrong

Answer 23

sorry I'm terrible at this :/

Answer 24

derivative is correct, plugging in might not be correct

Answer 25

oh and as i told its x^2, not x

Answer 26

so x^2 +16 = 9 +16 = 25
not -3+16

Answer 27

\[f(x)= 5- \frac{ 3 }{ 5} (x+3)\]

Answer 28

yes, there you go!

Answer 29

thank you so much! I was just making silly math errors.

Answer 30

just a moment

Answer 31

why is there a negative ?

Answer 32

a=-3, plugging into \[\frac{ x }{ \sqrt{x ^{2}-16} }\] gives you -3/5

Answer 33

sorry :P
thats correct

Answer 34

:)