Cho hàm số \(y = {x^3} - 3x + 2\). Các mệnh đề sau đây là đúng hay sai?

a) \(\mathop {\min }\limits_{\left[ {0\,;\,1} \right]} y = 0\).

b) \(\mathop {\min }\limits_{\left[ {0\,;\,2} \right]} y = y\left( 0 \right)\).

c) \(\mathop {\min }\limits_{\left[ { - 1\,;\,0} \right]} y + \mathop {\max }\limits_{\left[ {0\,;\,1} \right]} y = 4\).

d) \(\mathop {\min }\limits_{\left[ { - \frac{3}{2};0} \right]} \frac{1}{y} = \frac{8}{{25}}\).

a) Ta có  \(y' = 3{x^2} - 3\);   \(y' = 0 \Leftrightarrow x =  \pm 1\).

\(f\left( 0 \right) = 2;f\left( 1 \right) = 0\). Vậy \(\mathop {\min }\limits_{\left[ {0\,;\,1} \right]} y = 0\).

Chọn ĐÚNG.

b) Ta có \(y' = 3{x^2} - 3\); \(y' = 0 \Leftrightarrow x =  \pm 1\).

\(f\left( 0 \right) = 2;f\left( 1 \right) = 0;f\left( 2 \right) = 4\). Vậy \(\mathop {\min }\limits_{\left[ {0\,;\,2} \right]} y = f\left( 1 \right)\).

Chọn SAI.

c) Ta có \(y' = 3{x^2} - 3\); \(y' = 0 \Leftrightarrow x =  \pm 1\).

\(f\left( 0 \right) = 2;f\left( 1 \right) = 0;f\left( { - 1} \right) = 4\). Suy ra \(\mathop {\min }\limits_{\left[ { - 1\,;\,0} \right]} y = 2;\mathop {\max }\limits_{\left[ {0\,;\,1} \right]} y = 2\). Vậy \(\mathop {\min }\limits_{\left[ { - 1\,;\,0} \right]} y + \mathop {\max }\limits_{\left[ {0\,;\,1} \right]} y = 4\).

Chọn ĐÚNG.

d) Xét hàm số \(g\left( x \right) = \frac{1}{y} = \frac{1}{{{x^3} - 3x + 2}}\) trên \(\left[ {\frac{{ - 3}}{2};0} \right]\), ta có:

\(g'\left( x \right) = \frac{{3{x^2} - 3}}{{{{\left( {{x^3} - 3x + 2} \right)}^2}}}\); \(g'\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x =  - 1\\x = 1\end{array} \right.\).

\(g\left( { - 1} \right) = \frac{1}{4}\); \(g\left( { - \frac{3}{2}} \right) = \frac{8}{{25}}\); \(g\left( 0 \right) = \frac{1}{2}\).

Vậy \(\mathop {\min }\limits_{\left[ { - \frac{3}{2};0} \right]} \frac{1}{y} = \frac{1}{4}\).

Chọn SAI.