Question

f(x)=(7x)/((1+5x^2)^2) ,at an interval -5 is less then or equal to x, and x is less than or equal to 2.

The absolute maximum value is _____ and this occurs at x equals _____.
The absolute minimum value is _____ and this occurs at x equals _____.

Answer 1

you took the derivative?

Answer 2

I honestly don know how to take the derivative of the function.
\[f(x)=\frac{ 7x }{ (1+5x^2)^2 }\]

Answer 3

what is the derivative of the top?

Answer 4

7?

Answer 5

bottom?

Answer 6

2(1+5x^2)*(10x)?

Answer 7

@zzr0ck3r

Answer 8

yep

Answer 9

ok

Answer 10

derivative of top *bottom - derivative of bottom * top all over the bottom squared

Answer 11

sorry on tablet this is hard

Answer 12

\[\frac{7
*((1+5x^2)^2)-7x*2(1+5x^2)*(10x)?}{(((1+5x^2)^2)^2}\]

Answer 13

disregard the ?

Answer 14

\[\frac { (7)((1+5x^2)^2)-(2(1+5^2)^2)(10x)(7x) } { (1+5x^2)^2)^2 }\]

Answer 15

correct

Answer 16

oh yeah that's what I got hah

Answer 17

ok now we are setting this thing = 0 so we can disregard the bottom

Answer 18

so simplify the top

Answer 19

wait wait

Answer 20

there is a problem with the bottom derivative

Answer 21

((1+5x^2)^2
so
2(1+5x^2)*10x

Answer 22

we loose the ^2 on the outside

Answer 23

ohh. okay.

Answer 24

ok so now what do you have on top?

Answer 25

\[(7)((1+5x^2)^2)-(2(1+5x^2))(10x)(7x)\]

Answer 26

bah im sorry man its so hard to see on this thing your derivative is fine

Answer 27

...

Answer 28

ok the problem is we need to set this thing to = 0 and this is a 4th degree polynomial

Answer 29

7∗((1+5x 2 )^2−7x∗2(1+5x 2 )∗(10x) = 0 is hard to solve

Answer 30

I was just going to say the same.

Answer 31

factor 1+5x^2 out
(1+5x^2)(7(1+5x^2)-140x^2)=0

Answer 32

solve
5x^2+1=0
and this solution is complex so we disregard it

Answer 33

these solutions are complex*...

Answer 34

So what do we do now?

Answer 35

(7(1+5x^2)-140x^2) = 0
7+35x^2-140x^2 = 0
7-105x^2=0
x= +- sqrt(7/105)

Answer 36

so how do i find the max and min.and when it occurs as x?

Answer 37

one of those is the max and one is the min
so
take third derivative to find concavity (this is not going to be fun) or you can plug in numbers right next to those to tell
sqrt(7/105) is about .25
so plug in .24, sqrt(7/105) and .26
you will get some value for sqrt(7/105) and if the two other numbers are lower, then sqrt(7/105) is a max, and vice versa

Answer 38

or do you want to take the second derivative?

Answer 39

that should say second derivative two comments ago (not third)

Answer 40

you could also graph it and look at it:)
http://www.wolframalpha.com/input/?i=graph+%287x%29%2F%28%281%2B5x%5E2%29%5E2%29+

Answer 41

how do you want to do this @pdd21

Answer 42

I want to figure out the answer to the question the easiest way possible (: lol @zzr0ck3r

Answer 43

well, do you have to show work?

Answer 44

if you understand my argument concerning taking the points .24, sqrt(7/105), .26
then use that. its valid because we know that we only have two real places where slope = 0

Answer 45

no I dont it's online hw. I plug in an answer and it tells me if i'm from or right. so far I've tried this problem 12 times and still none correct -.-

Answer 46

what do you get when you evaluate the original function at sqrt(7/105)?

Answer 47

I got the max! (:
just the min
would it be the same values but neg?

Answer 48

no

Answer 49

sec

Answer 50

ps for the future
http://www.wolframalpha.com/input/?i=extrema+%287*x%29%2F%281%2B5x%5E2%29%5E2

Answer 51

ohhhh okay. lol and thanksss so much! (: I really appreciate you helping me! (: @zzr0ck3r

Answer 52

np, it looks like something happened in our setting = 0, but im sure you can figure it out. our x is off a little...but it gives the answer.