Question 40: Given the function y=f(x) , the variation table of the function y=f'(x) is as follows: The number of extreme points of the function \(y=f\left(x^{2}- 2 x\right)\) is

We have \(y^{\prime}=2(x-1) \cdot f^{\prime}\left(x^{2}-2 x\right)\)

\(y^{\prime}=0 \Leftrightarrow\left[\begin{array}{l}x=1\\f^{\prime}\left(x^{2}-2x\right)=0\end{array}\right\)\(\Leftrightarrow\left[\begin{array}{l}x=1\\x^{2}-2x=a\in(-\infty;-1)\\x^{2}-2x=b\in(-1;0)\\x^{2}-2x=c\in(0;1)\\x^{2}-2x=d\in(1;+\infty)\end{array}\right\)\(\Leftrightarrow\left[\begin{array}{l}x=1\\x^{2}-2xa=0a\in(-\infty;-1)\\\(1)\\x^{2}-2xb=0b\in(-1;0)\\\\\\\\\\\\(2)\\x^{2}-2xc=0c\in(0;1)\\\\\\\\\\\\\\\\\(3)\\x^{2}-2xd=0d\in(1;+\infty)\\\\\\\\\\(4)\end{array}\right\)

Equation (1) has no solution, equations (2),(3),(4) all have two distinct solutions other than 1, and because b, c, and d are different, the solutions of equation (2) ,(3),(4) are also different from each other.

Therefore \(f^{\prime}\left(x^{2}-2 x\right)=0\) There are 6 distinct solutions.

So y’=0 has 7 distinct solutions, so the number of extreme points of the function \(y=f\left(x^{2}-2 x\right)\) is 7

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