concavity, inflection points: y = 4/3x^3-2x^2+x-5

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tinad

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Jun 4, 2006
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Can someone help with the outcome of this function f(x)= 4/3x^3-2x^2+x-5
Fractions with powers is where i have no clue where to start. can someone show me how or explain why I should at a certain point?
 
f(x) = (4/3)x^3 - 2x^2 + x - 5

just use the power rule for derivatives ...

f'(x) = 4x^2 - 4x + 1

note that 3(4/3) = 4


a piece of advice ... get over the "fraction" anxiety now. you will be exposed to many more instances of dealing with fractions as you progress through the calculus.
 
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