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I don't believe that it's possible for a function to not have a local maxes and minimums. I have always been told that if you set the first derivative of a function to zero and solve for x and get imaginary numbers for your critical values that there are none. Now I don't by this because you can just keep on solving and get very defined values for critical points that are in the set of complex numbers... can maybe someone explain to me why it's considered not to have any critical points when you get complex numbers because by the looks of it every function has them they are just complex... like you can't deny that you can go through the same process to solve for critical points for equations that have them for equations that "don't" to get very defined values that are just complex... ? If they aren't critical points then what are they???

Just thinking about it a function MUST have a point were it turns around because if it didn't then it would just go on forever in one direction and never come back down...

For example the equation f(x) = (3X-1)/(2x^2 + x - 6) MUST have critical points were the function "turns around" because if it didn't it wouldn't be able to go back down... like when

f(0) = 1/6 f(1/2) = -.1 the function is decreasing and eventually becomes negative for values greater than 1/3... hmm so if there were no critical values then wouldn't you agree the function would just keep on going, becoming more and more negative as x becomes greater than 1/3... hmm but this isn't true as clearly it does turn around some were and become positive, for example f(3) = 8/15... how could f(x) become positive if there was no point were it turned around if it was just suppose to become more and more negative for values greater than 1/3???

For example setting the first derivative for this function equal to zero

x = 1/6 (2-7 i sqrt(2))

x = 1/6 (2+7 i sqrt(2))

so they are complex so there are "no" critical points... I'm not buying it... It seems as if these critical points are points for f(x) = x/0... now why is this? These seem to be the turning points... f(-2) and f(3/2) the y asymptote... hmmm

it's interesting that when you solve the first derivative when setting it equal to zero you only set the numerator and ignore the denominator completely because its "can't be defined", which I don't buy either but that is another issue...

sorry I'm just fascinated that people tell me that that f(x) = (3X-1)/(2x^2 + x - 6) has no critical points but yet it would appear that you can solve perfectly fine and get very well defined points... hope it makes sense can somebody explain why "there are none" as I'm not buying it... just thinking about it from a very logical perspective that there MUST be critical points to go from becoming more and more negative, if it's not suppose to turn around anywhere, but yet it becomes positive...? If there weren't any critical points wouldn't it just stay negative? hmmm....

Thanks!

Please tell me the truth please =)... no lies like cos(x)=-2 has no solutions lol...

2. Relevant equations

3. The attempt at a solution

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# Homework Help: I don't buy that it's possible for a function to not have critical points

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