Cho \(f\left(x\right)=\sin3x.\) Tính \(f"\left(-\dfrac{\pi}{2}\right)+f"\left(0\right)+f"\left(\dfrac{\pi}{18}\right)\).
\(-9-\dfrac{9\sqrt{3}}{2}\).\(-9+\dfrac{9\sqrt{3}}{2}\).\(4,5\).\(-13,5\).Hướng dẫn giải:\(f\left(x\right)=\sin3x\Rightarrow f'\left(x\right)=3\cos3x\Rightarrow f"\left(x\right)=-9\sin3x.\) Do đó \(f"\left(0\right)=0;f"\left(-\frac{\pi}{2}\right)=-9;f'\left(\frac{\pi}{18}\right)=-\frac{9}{2}\) và \(f"\left(-\frac{\pi}{2}\right)+f"\left(0\right)+f"\left(\frac{\pi}{18}\right)=-13,5\)