Đáp án:
$\underset{\left[-\dfrac{\pi}{2};\dfrac{\pi}{2}\right]}{max f(x) }=1$
Giải thích các bước giải:
$ y=3\sin x-4\sin^3x\\ y'=3\cos x-12\sin^2 x.(\sin x)'\\ =3\cos x-12\sin^2 x.\cos x\\ =3\cos x(1-4\sin^2 x)\\ y'=0\\ \Leftrightarrow \left[\begin{array}{l} \cos x=0\\ 1-4\sin^2 x=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} x=\dfrac{\pi}{2}+k\pi &(k\in \mathbb{Z}\\ \sin x=\pm\dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} x=\dfrac{\pi}{2}+k\pi &(k\in \mathbb{Z})\\ x=-\dfrac{\pi}{6}+l2\pi&(l\in \mathbb{Z})\\ x=\dfrac{\pi}{6}+m2\pi&(m\in \mathbb{Z})\end{array} \right.\\ x \in \left[-\dfrac{\pi}{2};\dfrac{\pi}{2}\right]\\ \Rightarrow x \in \left\{-\dfrac{\pi}{2};\dfrac{\pi}{2};-\dfrac{\pi}{6};\dfrac{\pi}{6}\right\}\\ f\left(-\dfrac{\pi}{2}\right)=1\\ f\left(\dfrac{\pi}{2}\right)=-1\\ f\left(-\dfrac{\pi}{6}\right)=-1\\ f\left(\dfrac{\pi}{6}\right)=1$
Vậy $\underset{\left[-\dfrac{\pi}{2};\dfrac{\pi}{2}\right]}{max f(x) }=1$
