Question

The linear approximation at x = 0 to f(x) = \sqrt { 8 + 2 x } is y =

Answer 1

\[\sqrt{8+2x}\]

Answer 2

x−−−−−

Its supposed to be the square root sign

\[\sqrt{8+2x}\]

Answer 3

yes

Answer 4

Got it.

Answer 5

how

Answer 6

Can someone help me with my question please
http://openstudy.com/study#/updates/545aecf7e4b017f3f584ed4b

Answer 7

Use the same formula from last time:
\[y=f(0)+f'(0)(x-0)\]

Answer 8

so its 0?

Answer 9

what's 0?

Answer 10

\[f'(x)=?\]

Answer 11

i mean sqrt of 8

Answer 12

well f(0) is sqrt(8)

Answer 13

so the derivative of that is 2?

Answer 14

Yes.

Answer 15

\[y=f(0)+f'(0)(x-0) \\ y=\sqrt{8}+f'(0)(x-0) \\ y=\sqrt{8}+f'(0)x\]
The derivative of sqrt(8+2x) is not 2.

Answer 16

Oh wait no its not 2

Answer 17

\[(8+2x)^\frac{ 1 }{ 2 }\] do this :)

Answer 18

its (2x+8)^-1/2

Answer 19

no manhani that negative exponent would make it a recipricol

Answer 20

wait what?

Answer 21

thats the derivative of the equation tho

Answer 22

XD yes yes sorry

Answer 23

but when i plug in sqrt 8 + the derivative its saying its the wrong answer

Answer 24

+ the derivative TIMES x

Answer 25

did you forget the final x?

Answer 26

2.82+x((2x+8)^(-1/2)) its not going through

Answer 27

which site are you using, sometimes they want it simplified a certain way

Answer 28

webwork its for my homework

Answer 29

The equation of the tangent line to f(x) at x = a is
L(x) = f(a) + f '(a) * ( x - a )

Answer 30

\[\sqrt{8}+\frac{ x }{ \sqrt{8+2x} }\]

Answer 31

y = what is above

Answer 32

try that and tell me what happens

Answer 33

incorrect

Answer 34

double check the equation you gave us then

Answer 35

The linear approximation at x = 0 to f(x) = \sqrt { 8 + 2 x } is y =

Answer 36

did you multiply x?

Answer 37

im stupid.... PLUG 0 INTO OUR DERIVATIVE!!

Answer 38

(2*0+8)^-1/2

Answer 39

i finally got it

Answer 40

Thank you for your time!!

Answer 41

:)

Answer 42

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Sorry if its really long.