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| The absolute maximum value of \(y=x^{3}-3x+2\) in \(0\leq x\leq 2\) is |
\(4\) \(6\) \(2\) \(0\) |
| \(4\) |
| \(\begin{aligned}y^{\prime}(x)&=3x^{2}-3\\ \text{So }y^{\prime}(x)&=0\Rightarrow x=\pm 1\\ y^{\prime\prime}(x)&=6x\\ y^{\prime\prime}(-1)&=-6<0\\ y(-1)&=-1+3+2=4\end{aligned}\) |
