Question

find critical points of (x+1)/(x^(2)+1)

Answer 1

i got the derivative as [(x^(2)+1)-(x+1)(2x)]/[(x^(2)+1]^2

Answer 2

Other than starting a new topic because you didn't like my answer... did you do that again?

And would you like me to do it step by step?

Answer 3

i changed the derivative because you said it wasn't right. you should probably check before accusing someone.

Answer 4

You started a new topic. It made me feel sad.

Give me a second. This has a lot of little stuffs.

Answer 5

the derivative is right.

Answer 6

what should i do next?

Answer 7

I've totally messed this question up. I'm sorry. I'm leaving it to someone elses capable hands.

Answer 8

find the roots of each side of the fraction. the roots are your critical numbers.

Answer 9

you mean solve for x? like (x^(2)+1)-(x+1)(2x)=0?

Answer 10

i tried that but got with x^2 = 2x-1

Answer 11

* got stuck

Answer 12

Oh. I can do that part. x^2=2x-1
x^2-2x+1
(x-1)(x-1)
x=1

Answer 13

that looks like a quadratic equation, where quadratic formula comes handy.

Answer 14

so 1 is the critical point? the only critical point?

Answer 15

the numerator is equivalent to this
\[\huge x^2+1=(x+1)(2x)\]

Answer 16

did you mean the denominator?

Answer 17

the numerator. \[\large (x^2+1)-(x+1)(2x) = 0\]

Answer 18

yes. and then if you it for x you get (x-1)(x-1) using quadratics. so 1 is a critical point?

Answer 19

i think the quadratic equation is \[\huge -x^2 \large - 2x + 1=0\]

Answer 20

yup. you are right. i missed a negative. now i get -2.414 and .4142

Answer 21

that's \[\large -1 \pm \sqrt{2} \] all right.

Answer 22

all right

Answer 23

so would those be the critical points?

Answer 24

just the two of them. the denominator yield no critical number.

Answer 25

ok. thank you

Answer 26

yw; and thank your for the medal.

Answer 27

@r0n4ld i was just checking the answers and i plugged -1 and sqrt(2) in the function. i got 0 if i plugged in -1 but sqrt(2)

Answer 28

*but not sqrt(2)

Answer 29

your roots in decimal form is the approximation to the exact roots, that is
\[ \large -2.414 \approx -1 - \sqrt{2}\] and \[\large 0.4142 \approx -1 + \sqrt{2} \]

Answer 30

so you mean that if i plug in -2.414 and .04142 then it'll be 0 just to confirm.

Answer 31

yes, the answer is zero in \[\huge f \prime(x)\] NOT \[\huge f(x)\]

Answer 32

oh. ok. i'll check it again then