Đáp án:
$\begin{cases}min[f(x)] = \dfrac{1}{2} \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\max[f(x)] = 1 \Leftrightarrow x = k\dfrac{\pi}{2}\end{cases}\qquad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}f(x) = \sin^4x + \cos^4x\\ = (\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x\\ = 1 - \dfrac{1}{2}\sin^22x\\ = 1 - \dfrac{1}{4}(1 - \cos4x)\\ = \dfrac{\cos4x + 3}{4}\\Ta\,\,có:\\ -1 \leq \cos4x \leq 1\\ \Leftrightarrow 2 \leq \cos4x + 3 \leq 4\\ \Leftrightarrow \dfrac{1}{2} \leq \dfrac{\cos4x + 3}{4} \leq 1\\ Hay \,\,\dfrac{1}{2} \leq f(x) \leq 1\\ Vậy\,\,min[f(x)] = \dfrac{1}{2} \Leftrightarrow \cos4x = -1 \Leftrightarrow x = \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\ max[f(x)] = 1 \Leftrightarrow \cos4x = 1 \Leftrightarrow x = k\dfrac{\pi}{2}\qquad (k \in \Bbb Z)\end{array}$
