Question 13: Given the function \(y=x^{3}-6 x^{2}+4 x-7 . \) . Let the coordinates of the two extreme points of the graph of the function be \(x_{1}, x_{2}\) . Then the value of the sum \(x_{1}+x_{2}\) is:
\(\begin{array}{l} y^{\prime}=3 x^{2}-12 x+4 \\ y^{\prime}=0 \Leftrightarrow 3 x^{2}-12 x+ 4=0 \end{array}\)
\(x_{1}, x_{2} \text { are two solutions of the equation } y^{\prime}=0 \text { . }\)
\(\text { Then, according to Viet’s theorem, we have: } x_{1}+x_{2}=4 \text { . }\)
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