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