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ksandoval
Group Title
how do i find the critical points of f(x) = (x2)^5 (x+3)^4?
 one year ago
 one year ago
ksandoval Group Title
how do i find the critical points of f(x) = (x2)^5 (x+3)^4?
 one year ago
 one year ago

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freewilly922 Group TitleBest ResponseYou've already chosen the best response.0
How would you explain a critical point? A minimum/maximum? a zero?
 one year ago

ksandoval Group TitleBest ResponseYou've already chosen the best response.0
hmm i dont know haha... i know that i need to find the derivative first though.
 one year ago

freewilly922 Group TitleBest ResponseYou've already chosen the best response.0
A 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,
 one year ago

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

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

ksandoval Group TitleBest ResponseYou've already chosen the best response.0
embarrassing, but i honestly have no idea how to factor that. ha.
 one year ago

freewilly922 Group TitleBest ResponseYou've already chosen the best response.0
Well, the idea is to find an alternative to brute force.
 one year ago

freewilly922 Group TitleBest ResponseYou've already chosen the best response.0
Ok, 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.
 one year ago
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