Question

Calc help . Find the absolute extrema of the fn over the closed interval. f(x)=x/(sqr.rt of x^2+1) [0, 2]

Answer 1

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

Answer 2

yes!

Answer 3

did you take the derivative?

Answer 4

yes thats what i did

Answer 5

did you get
\[f'(x)=\frac{1}{\sqrt{x^2+1}^3}\]?

Answer 6

nope :[

Answer 7

oh too bad

Answer 8

perhaps you got a big fat mess that looked hard to simplify

Answer 9

in the end i got x^2+1/(x^2+1)^1/2

Answer 10

you need the quotient rule for this one

Answer 11

i think so :( hey that was mean :`(

Answer 12

that wat i did

Answer 13

ok lets do that second, first lets answer the question
the derivative is always positive, which means the function is always increasing
therefore the min is at the left endpoint, namely at \(x=0\) and the max is at \(x=2\)

Answer 14

ok :l

Answer 15

for the derivative you need
\[\frac{gf'-fg'}{g^2}\] with \[f(x)=x, f'(x)=1,g(x)=\sqrt{x^2+1}, g'(x)=\frac{x}{\sqrt{x^2+1}}\]

Answer 16

ok

Answer 17

you get by direct substitution
\[\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}\]

Answer 18

huh? whatha :`(

Answer 19

now to simplify this ugly mess, multiply top and bottom by \(\sqrt{x^2+1}\) to clear the compound fraction

Answer 20

ok

Answer 21

uh oh maybe we should go back and make sure it is clear how to find the derivative using the quotient rule, unless what i wrote was clear
i sense it was not

Answer 22

wait ok dont u h=get -2x^2

Answer 23

no i get it its just that i got -2x^2 instead of -x^2 :l

Answer 24

no if you multiply top and bottom by \(\sqrt{x^2+1}\) the numerator becomes
\[x^2+1-x^2\]

Answer 25

i mean for the numerator i got okay i got (x^2+1)^1/2-x^2/2(x^2+1)^1/2 all over x^2+1

Answer 26

\[\left(\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}\right)\times \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}\]
\[=\frac{x^2+1-x^2}{(x^2+1)^{\frac{3}{2}}}\]

Answer 27

yes what you wrote is correct

Answer 28

okay :}

Answer 29

this is right
(x^2+1)^1/2-x^2/2(x^2+1)^1/2 all over x^2+1

Answer 30

so then.....where du i go from what i had multiply by (x^2+1)^1/2??

Answer 31

but in the end, after getting rid of the compound fraction you are left with
\[f'(x)=\frac{1}{(x^2+1)^{\frac{3}{2}}}\]

Answer 32

yes that is the easiest way to do it

Answer 33

o okay :]

Answer 34

the radical is gone from the numerator, leaving only 1 up top
since the numerator is 1, it is never zero, so there are no critical points
the derivative is always positive
function is always increasing
here is a nice picture
http://www.wolframalpha.com/input/?i= \frac{x}{\sqrt{x^2%2B1}}+domain+0..2

Answer 35

link didn't work but if you copy and paste you will see it

Answer 36

ok i sorta got it thanks!

Answer 37

like i said (not trying to be mean, really) it is the messy algebra that is a pain
only reason i know it is because i have seen it several times