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freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0How would you explain a critical point? A minimum/maximum? a zero?

ksandoval
 2 years ago
Best ResponseYou've already chosen the best response.0hmm i dont know haha... i know that i need to find the derivative first though.

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0A critical point of a function of a single real variable, ƒ(x), is a value x0 in the domain of ƒ where either the function is not differentiable or its derivative is 0,

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0So it sounds like you need to take the derivative, set it to zero and then solve for x.

ksandoval
 2 years ago
Best ResponseYou've already chosen the best response.0and i got \[f'(x) = 5(x2)^4(x+3)^4 + 4(x+3)^3(x2)^5\]

ksandoval
 2 years ago
Best ResponseYou've already chosen the best response.0embarrassing, but i honestly have no idea how to factor that. ha.

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0Well, the idea is to find an alternative to brute force.

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, when things look ugly like this, substitute something easy. Like A= x2 and B=x+3 This gives you \[f'(X) = 5A^4B^4 +4B^3A^5\] This is easy to factor and set to zero. Then once you have solved for when THIS is zero with respect to A and B then resubstitute the A=x2 and B=x+3 to get you final answer.
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