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anonymous
 3 years ago
help! :( finding the critical point of g(x)= (x+4)/(x1)
anonymous
 3 years ago
help! :( finding the critical point of g(x)= (x+4)/(x1)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i used the quotient rule!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i got my critical point 5 and 1 but it's wrong :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i doubt there are any

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the derivative is \(\frac{5}{(x1)^2}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a fraction is only zero if the numerator is zero this numerator is \(5\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right so when i get to this: 5/(x^22x+1)= 0 what do i do?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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
 3 years ago
Best ResponseYou've already chosen the best response.0you may remember what such a thing looks like from a pre  calc course

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok but then why isn't 5 considered a critical point?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know 1 gives you undefined.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a 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
 3 years ago
Best ResponseYou've already chosen the best response.0you mean the denominator?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0graph 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
 3 years ago
Best ResponseYou've already chosen the best response.0i did, but what i don't understand why it has no critical point.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay but in this question im not allowed to use the calculator.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well, because it doesn't does \(f(x)=2x+1\) have a critical point?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so if i can't graph it, how do can i automatically tell there is no critical point by looking at the derivative 5/(x^22x+1)= 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0of course it doesn't, because for \(f(x)=2x+1\)you have a line that is always increasing

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when is a fraction equal to zero?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and 0 does not multiply 5..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is \(\frac{5}{x}=0\) solvable?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right, and neither is \(\frac{5}{(x1)^2}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks @satellite73 :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how about \(e^x\)? is that ever zero? yw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no, \(e^x>0\) for all \(x\) \(e^1=e\) not zero

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the function \(f(x)=e^x\) is strictly increasing.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@satellite73 i have a question: how come 5/(x1)=0 is not a critical point and 4/15 is?
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