Question

Find the linearization of f(x) =1/sqrt(x+1) at x=0

Answer 1

you know how find tangent lines right
this is what it is asking you to do
find the tangent line of the curve at x=0

Answer 2

if x=0, then y=1/sqrt{0+1}=1/1=1
so we know a point on the the tangent line which is (1,1)
find f' and plug in 0 for x to find the slope of the tangent line
so then you can plug everything you find into
y-y_1=m(x-x_1)

Answer 3

I got y=1/10sqrt(5)+1/5, but that not one of the answer choices

Answer 4

I got y=1/10sqrt(5)x+1/5, but that not one of the answer choices

Answer 5

\[f'(x)=\frac{-1}{2}(x+1)^\frac{-3}{2}=\frac{-1}{2(x+1)^\frac{3}{2}} =>f'(0)=\frac{-1}{2}\]

Answer 6

\[y-1=\frac{-1}{2}(x-0)\]

Answer 7

it should be x+5 in the parenthesis instead of x+1

Answer 8

ok well i did it for y=1/sqrt{x+1} (<-what you wrote above?)
but you can do it for y=1/sqrt{x+5}
use what i did as an example

Answer 9

I wrote the problem wrong it should be x+5 inside the square root, sorry!

Answer 10

y-1/5=1/10sqrt(5)(x-0)

Answer 11

what happen to the square root when you plugged in 0 for x?

Answer 12

the 5 is still inside the square root

Answer 13

\[f(0)=\frac{1}{\sqrt{0+5}}=\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}}{5}\]

Answer 14

\[y-f(0)=m(x-0)\]

Answer 15

so we need m

Answer 16

\[f'(x)=\frac{-1}{2}(x+5)^\frac{-3}{2}=\frac{-1}{2(x+5)^\frac{3}{2}} =>f'(0)=\frac{-1}{2(5)^\frac{3}{2}}\]

Answer 17

m=f'(0)

Answer 18

here are the answers choices

Answer 19

it doesn't match the answer choices

Answer 20

i see the answer

Answer 21

try simplifying and writing it
as y=mx+b
by adding f(0) on both sides

Answer 22

hello all
@ta i wrote out an answer for your last problem

Answer 23

in your previous post
oh actually it was this problem! i wrote it out, it is probably still there

Answer 24

write as y=b+mx
then factor anything the b and m have in common and voolah

Answer 25

of course i made my usual typo, but i fixed it