anonymous
  • anonymous
help! :( finding the critical point of g(x)= (x+4)/(x-1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i used the quotient rule!
anonymous
  • anonymous
i got my critical point 5 and 1 but it's wrong :(
anonymous
  • anonymous
-5/(x^2-2x+1)= 0

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

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