\(f\left(x\right)=x^6+4\left(1-x^2\right)^3\)
=>\(f'\left(x\right)=6x^5+4\cdot3\cdot\left(1-x^2\right)^2\cdot\left(1-x^2\right)'\)
=>\(f'\left(x\right)=6x^5+12\left(-2x\right)\left(x^2-1\right)^2\)
=>\(f'\left(x\right)=6x^5-24x\left(x^4-2x^2+1\right)\)
=>\(f'\left(x\right)=6x^5-24x^5+48x^3-24x=-18x^5+48x^3-24x\)
Đặt f'(x)=0
=>\(x\left(-18x^4+48x^2-24\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\18x^4-48x^2+24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=2\\x^2=\dfrac{2}{3}\end{matrix}\right.\)
mà -1<=x<=1
nên \(x\in\left\{0;-\sqrt{\dfrac{2}{3}};\sqrt{\dfrac{2}{3}}\right\}\)
\(f\left(-1\right)=\left(-1\right)^6+4\left[1-\left(-1\right)^2\right]^3=1\)
\(f\left(1\right)=1^6+4\left(1-1^2\right)^3=1\)
\(f\left(0\right)=0^6+4\left(1-0^2\right)^3=4\)
\(f\left(-\sqrt{\dfrac{2}{3}}\right)=\left(-\sqrt{\dfrac{2}{3}}\right)^6+4\cdot\left[1-\left(-\sqrt{\dfrac{2}{3}}\right)^2\right]^3=\dfrac{4}{9}\)
\(f\left(\sqrt{\dfrac{2}{3}}\right)=\left(\sqrt{\dfrac{2}{3}}\right)^6+4\cdot\left[1-\left(\sqrt{\dfrac{2}{3}}\right)^2\right]^3=\dfrac{4}{9}\)
Do đó, \(f\left(x\right)_{min\left[-1;1\right]}=\dfrac{4}{9};f\left(x\right)_{max\left[-1;1\right]}=4\)
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