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  • 5 years ago

critical points of .34x3 + 6.95x2 + 46.56x + 101.362

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  1. anonymous
    • 5 years ago
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    first call your function f(x), so you'll have: \[f(x) = 0.34x^3 + 6.95x^2 + 46.56x + 101.362\] In order to find the critical points, you'll have to find the first derivative, then you can take 2 conditions to find the critical points which are: (a) f'(x) = 0 b) f'(x) DNE = Doesn't Exist, and it doesn't exist when you either have x in the denominator, or you have the sqrt of negative number. So in this case, your function is a polynomial and it's continuous everywhere, so after you find the first derivative, use the first condition to find the zeros of f'(x) , those zeros are your critical numbers, so to find your (y) for the critical numbers, subsitite the (x) that you'll find in the function and you'll get the critical points Do you understand the concept better now? ^_^

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