Question

help! :( finding the critical point of g(x)= (x+4)/(x-1)

Answer 1

i used the quotient rule!

Answer 2

i got my critical point 5 and 1 but it's wrong :(

Answer 3

-5/(x^2-2x+1)= 0

Answer 4

i doubt there are any

Answer 5

why?

Answer 6

the derivative is \(\frac{-5}{(x-1)^2}\)

Answer 7

a fraction is only zero if the numerator is zero
this numerator is \(-5\)

Answer 8

-5/(x^2-2x+1)= 0

Answer 9

right so when i get to this: -5/(x^2-2x+1)= 0 what do i do?

Answer 10

the denominator is positive for all values of \(x\) except for \(x=1\) where both it, and the original function are undefined
the numerator is negative
that tells you the function is always decreasing

Answer 11

right

Answer 12

you may remember what such a thing looks like from a pre - calc course

Answer 13

ok but then why isn't -5 considered a critical point?

Answer 14

i know 1 gives you undefined.

Answer 15

a critical point is not the numerator
a critical point is a number \(a\) where either \(f'(a)=0\) or \(f'(a)\) does not exist

Answer 16

you mean the denominator?

Answer 17

graph your derivative
you will see that it never crosses the \(y\) axis, that it always lives below it
i.e. it is always negative
if you graph your original function, you will see that it is always decreasing

Answer 18

i did, but what i don't understand why it has no critical point.

Answer 19

okay but in this question im not allowed to use the calculator.

Answer 20

well, because it doesn't
does \(f(x)=2x+1\) have a critical point?

Answer 21

so if i can't graph it, how do can i automatically tell there is no critical point by looking at the derivative -5/(x^2-2x+1)= 0

Answer 22

of course it doesn't, because for \(f(x)=2x+1\)you have a line that is always increasing

Answer 23

when is a fraction equal to zero?

Answer 24

oh no it doesn't.

Answer 25

and 0 does not multiply -5..

Answer 26

is \(\frac{5}{x}=0\) solvable?

Answer 27

nope.

Answer 28

i think i got it...

Answer 29

right, and neither is \(\frac{-5}{(x-1)^2}\)

Answer 30

thanks @satellite73 :)

Answer 31

how about \(e^x\)? is that ever zero?
yw

Answer 32

no

Answer 33

1?

Answer 34

no, \(e^x>0\) for all \(x\)
\(e^1=e\) not zero

Answer 35

the function \(f(x)=e^x\) is strictly increasing.

Answer 36

right.

Answer 37

@satellite73 i have a question: how come -5/(x-1)=0 is not a critical point and -4/15 is?