Question

calculus help

Answer 1

for \[f(x)=\frac{ 2x+2 }{ \sqrt{x^2}+1 }\], \[f'(x)=\frac{ 2-2x }{ (x^2 +1)^{3/2}}\]
what is the range of f(x)?

Answer 2

this must be the original function

\[\frac{2x+2}{\sqrt{x^2+1}}\]

Answer 3

yes @Zarkon

Answer 4

Graph it at desmosgraphingcalculator.com

Answer 5

from there you can determine the range

Answer 6

i have to figure it out without graphing it

Answer 7

o_o

Answer 8

that was my face also lol

Answer 9

From the derivative, I figure out you mean
\(\Large f(x)=\frac{ 2x+2 }{ \sqrt{x^2+1} }\).

We realize from f(x) that there is no vertical asymptote since the denominator is always \(\ge \)1.

As |x| gets larger, so does the denominator, therefore there will be two horizontal asymptotes by taking the limit of f(x) as x->\(\pm \infty\).

Also, locate the critical point where f'(x)=0, find out if it is a max. or min (using the second derivative test.

Answer 10

I think, the range is between 2 and 2*root(2)

Answer 11

how did you get that?

Answer 12

wait. i will send you

Answer 13

and you have to rotate it by 180 :)

Answer 14

\[\lim_{x\to-\infty}\frac{2x+2}{\sqrt{x^2+1}}=-2\]

Answer 15

@Zarkon no. it's not.

Answer 16

@Zarkon you can't just put values in limits. you need to make it in way that it give comprehensive answers.
would you be able to explain how you say it is -2

Answer 17

\[\frac{2x+2}{\sqrt{x^2+1}}=\frac{2x+2}{\sqrt{x^2(1+\frac{1}{x^2})}}=\frac{2x+2}{\sqrt{x^2}\sqrt{(1+\frac{1}{x^2})}}\]

\[=\frac{2x+2}{|x|\sqrt{(1+\frac{1}{x^2})}}=\frac{x}{|x|}\frac{2+\frac{2}{x}}{\sqrt{1+\frac{1}{x^2}}}\]

Answer 18

yes. that's exactly right. and it give +2 for both -infinity and +infinity of x

Answer 19

wrong

Answer 20

it is -2...think about it for a while

Answer 21

@Zarkon no way. well, I need some explainations on that. I still argue that it is +2

Answer 22

http://www.wolframalpha.com/input/?i=%282x%2B2%29%2Fsqrt%28x%5E2%2B1%29+x+to+-infinity

Answer 23

@Zarkon that website has issues. can you give reasonable answer please?

Answer 24

@Zarkon I mean, just think about for a second, You can't say it is -2.

your work on limits is correct. and absolute value doesn't matter in this case. So x is cancel out. then substitute -infinity for x.

as you can see. 1/-infinity is almost 0. so it doesn't matter either.

finally you get 2/toor(1). which is 2

Answer 25

I think you need to go back and review limits

Answer 26

\(\frac{x}{|x|}=-1\) when \(x<0\)

Answer 27

@Zarkon okay. May be I am wrong. can you point out my mistake?

Answer 28

\[\lim_{x\to-\infty}\frac{2x+2}{\sqrt{x^2+1}}=\lim_{x\to-\infty}\frac{x}{|x|}\frac{2+\frac{2}{x}}{\sqrt{1+\frac{1}{x^2}}}=(-1)\frac{2+0}{\sqrt{1+0}}=-2\]

Answer 29

@Zarkon My apologies. I am wrong. That's for that.

Answer 30

so what's the range then?

Answer 31

@Zarkon I can't make my mind.:) it has to be 2.:D

Answer 32

last thing I type here

\[\frac{2x+2}{\sqrt{x^2+1}}\]

\[x^2+1\ge1\]

so the denominator is always positive

\[2x+2<0\] if \[x<-1\]

thus \[\frac{2x+2}{\sqrt{x^2+1}}\]

has to be negative for x<1...and therefore its limit can't be +2

Answer 33

less than -1 not 1 (typo)

Answer 34

has to be negative for x<-1...and therefore its limit can't be +2

Answer 35

@Zarkon I agree with you. After long time. I decided to use online tools. so
https://www.desmos.com/calculator/22wqwgryta

rangeis [-2, 2root(2)]

Answer 36

\[(-2,2\sqrt{2}]\]

Answer 37

the function never reaches -2