Question 33: Given the function \(y = {x^3} – 2{x^2} – 1\) (1) and the propositions (1) The extreme point of the function (1) is x = 0 or x = 4/3 (2) The extreme point of function (1) is x = 0 and x = 4/3 (3) The extreme point of the graph of function (1) is x = 0 and x = 4/ 3 (4) The extrema of the function (1) is x = 0 and x = 4/3 In the above statements, the number of false statements is:

We have: \(y’ = 3{x^2} – 4x,y” = 6x – 4;\)

\(y’ = 0 \Leftrightarrow 3{x^2} – 4x = 0 \Leftrightarrow \left[\begin{array}{l}[\begin{array}{l}

x = 0\\

x = \frac{4}{3}

\end{array} \right.\)

y”(0) = -4 < 0; y''(4/3) = 4 > 0. So the function has two extremes, x = 0 and x = 4/3

The propositions (1); (2) and (3) are false; proposition (4) is true.

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