Question

What is the critical point of the function:
y=(a/x)+1

Answer 1

where a is a constant?

f'(x)=a(dy/dx[1/x])

\[a(\frac{-1}{x^2})\]

Answer 2

ya

Answer 3

\[=\frac{-a}{x^2}\]

Answer 4

when -a =0

Answer 5

oh whoops i forgot to say

Answer 6

?

Answer 7

that x has to be greater than 0

Answer 8

yes true because if it is 0 it's undefined. so a=0, x cannot equal 0

Answer 9

but I had to find the critical point

Answer 10

what is the x coordinate

Answer 11

the point depends on whether a is 0 or not

Answer 12

well lets say it isn't zero

Answer 13

there would be no critical

Answer 14

I don't think so

Answer 15

like graph the function with a equaling greater than zero and see that it doesn't make sense what u said

Answer 16

a or x is greater than zero?

Answer 17

A

Answer 18

if a equals zero than we are left with y=1

Answer 19

I don't know I am so confused

Answer 20

yes but the critical point happens when you have a horizontal tangent or when the derivative of the function is 0

Answer 21

maybe there is no critical point

Answer 22

what you have when x>0 is a saddle point

Answer 23

Sorry what si a saddle point?

Answer 24

because it gets as close to a 0 slope but never quite hits it

Answer 25

oh that is what i thought

Answer 26

so basically there is no critical point

Answer 27

when a = 0 you get a y=1 line correct? which has a slope of 0

Answer 28

actually you get y=x

Answer 29

y=(0/x)+1

y=1

Answer 30

a line which corsses the origin at 0,0

Answer 31

OMG i wrote the equation wrong

Answer 32

y = 1 doesn't cross at the origin y=1

Answer 33

K r u still interested in helping me out?

Answer 34

write it down i gotta study but do it quick

Answer 35

cuz i wrote the equation wrong

Answer 36

f(x)=a/x+x

Answer 37

i was going to say that's a horrible question because if a=0 then there is infinite critical points

Answer 38

f'(x)=(-a/x^2)+1

Answer 39

ya

Answer 40

oh x=a

Answer 41

0=(-a/x^2)+1
-1*(-x^2)=(-a/x^2)*-x^2 {multiplying by -x^2 to get rid of -

x^2=a

Answer 42

which would make sense since

\[0=\frac{-x^2}{x^2}+1=-1+1=0\]

Answer 43

so solving that you can find x easily

Answer 44

huh? how did u get rid of the a?

Answer 45

\[\pm \sqrt{a}\]

Answer 46

ya that is what i got

Answer 47

Thanks i really appreciate ur time

Answer 48

i just substituted x^2 = a to see if it cancelled which it did

Answer 49

no problem

Answer 50

oh ok

Answer 51

Don't waste ur time on here and go study!

Answer 52

haha yeah probably should but i usually stuyd at night so

Answer 53

I just find this site addicting. it is funner to solve others math problems than ur own

Answer 54

r u still there?
I have another question

Answer 55

whats up

Answer 56

Well I think I got it neways but u can sheck it out. only if u have the time

Answer 57

y=(1/x)+bx
Find the critical points

Answer 58

y'=-1/x^2 +b

Answer 59

x^2=(1/b)

Answer 60

x= +- sqrt (1/b)

Answer 61

thats correct

Answer 62

Thanks

Answer 63

if you don't think it is true put 1/b into the equation for x^2 youll get

-1/(1/b)+b =0

flip the (1/b)

-b+b=0

Answer 64

oh i get that now

Answer 65

hey wait it cld only be +sqrt(a) and + sqrt(1/b) since x has to be greater than 0

Answer 66

I need help. Are u still there?

Answer 67

Are u still there? I am stuck again

Answer 68

HELP?!?! Don't feel pressured whatsoever

Answer 69

what?

Answer 70

yeah it'd be >

Answer 71

ok so 4get it

Answer 72

I just neways missed my deadline I think I will just stay up all night Got too much work

Answer 73

missed your deadline? online class?

Answer 74

ya

Answer 75

It was at 1am and now it is 7 minutes later and I have loads of work to still hand in

Answer 76

When u r finished and u still want to help out give me a shout

Answer 77

I have one stupid question that I can't figure out!!!!