Question 21: The graph of the function y = 3×4 – 4×3 – 6×2 + 12x + 1 reaches its minimum at M(x1; y1). Sum x1 + y1

We have: y’ = 12x^{3} – 12x^{2 }– 12x + 12.

\(y’ = 0 \Leftrightarrow \left[\begin{array}{l}[\begin{array}{l}

x = 1 \Rightarrow y = 6\\

x = – 1 \Rightarrow y = – 10

\end{array} \right.\)

Variation table

⇒ M(-1;-10) x_{first} + y_{first} = -11

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