Question 30: Given a function y=f(x) determined on \(\mathbb{R}\) and a function y=f'(x) with a graph as shown in the figure. Find the number of extreme points of the function \(y=f\left(x^{2}-3\right)\)
Observe the graph we have \(y=f^{\prime}(x)\) change the sign from negative to positive through x=-2, so the function y=f(x) has an extreme point of x=-2 .
We have \(y^{\prime}=\left[f\left(x^{2}-3\right)\right]^{\prime}=2 x \cdot f^{\prime}\left(x^{2}-3\right)=0 \Leftrightarrow\left[\begin{array}{l}x=0\\x^{2}-3=-2\\x^{2}-3=1\end{array}\Leftrightarrow\left[\begin{array}{l}x=0\\x=\pm1\\x=\pm2\end{array}\right\right\)
But \(x=\pm 2\) is a double kamma, and the remaining solutions are single, so the function \(y=f\left(x^{2}-3\right)\) there are three extremes.
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