Đáp án:
$Max= \dfrac{3}{4}$
$min = \dfrac{1}{4}$
Giải thích các bước giải:
$y=x^6-3x^4+\dfrac{9}{4}x^2+\dfrac{1}{4}$
$y'=6x^5-12x^3+\dfrac{9}{2}x$
Giải phương trình $y'=0$
$\to x(6x^4-12x^3+\dfrac{9}{2})=0$
$\to \left[ \begin{array}{l} \left[ \begin{array}{l}x=0\\x=\dfrac{\sqrt{2}}{2} \end{array} \right.\\ \left[ \begin{array}{l}x=-\dfrac{\sqrt{2}}{2}\\ \left[ \begin{array}{l}x=\dfrac{\sqrt{6}}{2}\\x=-\dfrac{\sqrt{6}}{2} \end{array} \right.\end{array} \right. \end{array} \right.$
Thay :
$f(-1)= \dfrac{1}{2}$
$f(1)=\dfrac{1}{2}$
$f(0)= \dfrac{1}{4}$
$f(\dfrac{\sqrt{2}}{2})= \dfrac{3}{4}$
$f(-\dfrac{\sqrt{2}}{2})= \dfrac{3}{4}$
$f(\pm \dfrac{\sqrt{6}}{2})= \dfrac{1}{4}$
Vậy $Max= \dfrac{3}{4}$
$min = \dfrac{1}{4}$
