Question 48: The function y = x3 – 3x + 2 reaches its maximum at

We have: \(y’ = 3{x^2} – 3,y’ = 0 \Leftrightarrow \left[\begin{array}{l}[\begin{array}{l}
x = 1\\
x = – 1
\end{array} \right.\)

Consider the sign y’:

\(y’ > 0 \Leftrightarrow \left[\begin{array}{l}[\begin{array}{l}
x < - 1\\
x > 1
\end{array} \right.;y’ < 0 \Leftrightarrow - 1 < x < 1\)

Then we have a function that maximizes at x = -1.