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

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

Mathematics
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i used the quotient rule!
i got my critical point 5 and 1 but it's wrong :(
-5/(x^2-2x+1)= 0

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i doubt there are any
why?
the derivative is \(\frac{-5}{(x-1)^2}\)
a fraction is only zero if the numerator is zero this numerator is \(-5\)
-5/(x^2-2x+1)= 0
right so when i get to this: -5/(x^2-2x+1)= 0 what do i do?
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
right
you may remember what such a thing looks like from a pre - calc course
ok but then why isn't -5 considered a critical point?
i know 1 gives you undefined.
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
you mean the denominator?
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
i did, but what i don't understand why it has no critical point.
okay but in this question im not allowed to use the calculator.
well, because it doesn't does \(f(x)=2x+1\) have a critical point?
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
of course it doesn't, because for \(f(x)=2x+1\)you have a line that is always increasing
when is a fraction equal to zero?
oh no it doesn't.
and 0 does not multiply -5..
is \(\frac{5}{x}=0\) solvable?
nope.
i think i got it...
right, and neither is \(\frac{-5}{(x-1)^2}\)
thanks @satellite73 :)
how about \(e^x\)? is that ever zero? yw
no
1?
no, \(e^x>0\) for all \(x\) \(e^1=e\) not zero
the function \(f(x)=e^x\) is strictly increasing.
right.
@satellite73 i have a question: how come -5/(x-1)=0 is not a critical point and -4/15 is?

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