Question 43: Given the function \(y = f\left( x \right)\). The function \(y = f’\left( x \right)\) has a graph as shown. The number of extreme points of the function graph \(y = g\left( x \right) = f\left( {{x^2} – 4x + 3} \right) – 3{\left( {x – 2 } \right)^2} + \frac{1}{2}{\left( {x – 2} \right)^4}\) is

We have \(g’\left( x \right) = 2\left( {x – 2} \right)f’\left( {{x^2} – 4x + 3} \right) – 6\left( {x – 2} \right) + 2{\left( {x – 2} \right)^3}\)

\(g’\left( x \right) = 2\left( {x – 2} \right)\left[ {f’\left( {{x^2} – 4x + 3} \right) + {x^2} – 4x + 1} \right]\)

\(g’\left( x \right) = 0 \Leftrightarrow \left[\begin{array}{l}x=2\\f’\left({{x^2}–4x+3}\right)=2–\left({{x^2}–4x+3}\right)\end{array}\right\)[\begin{array}{l}x=2\f’\left({{x^2}–4x+3}\right)=2–\left({{x^2}–4x+3}\right)\end{array}\right\)

From the graph of the function

We have a line y = 2 – x that intersects the graph \(y = f’\left( x \right)\) at four distinct points whose coordinates are x = – 2;x = 0;x = 1;x = 2

So[\begin{array}{l}x=2\{x^2}–4x+3=–2\{x^2}–4x+3=0\{x^2}–4x+3=1\{x^2}–4x+3=2\end{array}\right\Leftrightarrow\left[\begin{array}{l}x=2\x=1\x=3\x=2\pm\sqrt2\x=2\pm\sqrt3\end{array}\right\)

\( \Leftrightarrow \left[\begin{array}{l}x=2\\{x^2}–4x+3=–2\\{x^2}–4x+3=0\\{x^)2}–4x+3=1\\{x^2}–4x+3=2\end{array}\right\Leftrightarrow\left[\begin{array}{l}x=2\\x=1\\x=3\\x=2\pm\sqrt2\\x=2\pm\sqrt3\end{array}\right\)

We have BBT:

From BBT, deduce the graph of the function with 6 extreme points.

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