Đáp án:
$\mathop{\max}\limits_{[-1;1]}f(x) = \dfrac34\Leftrightarrow x = \pm \dfrac{\sqrt2}{2}$
$\mathop{\min}\limits_{[-1;1]}f(x) =\dfrac14\Leftrightarrow x = 0$
Giải thích các bước giải:
$\quad y = f(x) =x^6 - 3x^4 + \dfrac{9}{4}x^2 + \dfrac14$
Đặt $t = x^2\quad (t\geqslant 0)$
$x \in [-1;1] \Rightarrow t \in [0;1]$
Ta được:
$\quad y = f(t) = t^3 - 3t^2 + \dfrac{9}{4}t + \dfrac{1}{4}$
$\Rightarrow y' = f'(t) = 3t^2 - 6t + \dfrac{9}{4}$
$f'(t) = 0 \Leftrightarrow \left[\begin{array}{l} t = \dfrac{1}{2}\\t = \dfrac{3}{2}\end{array}\right.$
Bảng xét dấu:
\(\begin{array}{c|ccc}
t&0&&\dfrac12&&1&&\dfrac32&&+\infty\\\hline
f'(t)&\vert&+&0&-&\vert&-&0&+&
\end{array}\)
Dựa vào bảng xét dấu ta được:
$\mathop{\max}\limits_{[0;1]}f(t) = f\left(\dfrac12\right) = \dfrac{3}{4}$
Ta lại có:
$f(0) = \dfrac14;\quad f(1) = \dfrac12$
$\Rightarrow \mathop{\min}\limits_{[0;1]}f(t) = f(0) = \dfrac14$
Do đó:
$\mathop{\max}\limits_{[-1;1]}f(x) = \dfrac34\Leftrightarrow t = \dfrac12 \Leftrightarrow x = \pm \dfrac{\sqrt2}{2}$
$\mathop{\min}\limits_{[-1;1]}f(x) =\dfrac14 \Leftrightarrow t = 0\Leftrightarrow x = 0$
