(1,5 points) Find the intervals of concavity and the inflection points of the function f(x)=x(2x−1)^2

Jaya Legge

Jaya Legge

Answered question

2021-06-12

(1,5 points) Find the intervals of concavity and the inflection points of the function f(x)=x(2x1)2

Answer & Explanation

bahaistag

bahaistag

Skilled2021-06-13Added 100 answers

Condition for concavity: f(x)>0 Condition for inflection point: f(x)=0
For the given function: f(x)=24x8
f(13)=24×138=0
For x>3,f(x) is greater than zero.
Therefore, (3, 75) point is the inflection point of f(x) and for x>3f(x) is concave.

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