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Question Number 77574 by jagoll last updated on 08/Jan/20

$$whatismaximum$$$$valueofy={(3x+4)}^2\times \sin \left(2x\right)$$

Commented by MJS last updated on 08/Jan/20

$$\sin 2x=-1\Rightarrow x=n\pi -\frac{\pi }{4}=\frac{4n-1}{4}\pi$$$$\sin 2x=+1\Rightarrow x=n\pi +\frac{\pi }{4}=\frac{4n+1}{4}\pi$$$$n\in \mathbb{Z}$$$$$$$$f\left(x\right)={(3x+4)}^2\sin 2x$$$$f\left(\frac{4n-1}{4}\pi \right)=-\frac{{(12\pi n-3\pi +16)}^2}{16}$$$$n=\pm {10}^3\Rightarrow f\left(x\right)\approx -8.9\times {10}^7$$$$n=\pm {10}^6\Rightarrow f\left(x\right)\approx -8.9\times {10}^{13}$$$$n=\pm {10}^9\Rightarrow f\left(x\right)\approx -8.9\times {10}^{19}$$$$...$$

Commented by jagoll last updated on 08/Jan/20

$$please$$

Commented by MJS last updated on 08/Jan/20

$$\mathrm{no}\mathrm{maximum}\mathrm{because}\underset{x\to \pm \mathrm{\infty }}{\lim }{(3x+4)}^2=+\mathrm{\infty }$$

Commented by jagoll last updated on 08/Jan/20

$$ifminimumsir?$$

Commented by MJS last updated on 08/Jan/20

$$\mathrm{also}\mathrm{no}\mathrm{minimum}$$$$-1⩽\sin 2x⩽1$$$$-\mathrm{\infty }<{(3x+4)}^2\sin 2x<+\mathrm{\infty }$$

Commented by jagoll last updated on 08/Jan/20

$$thanksyousir$$

Commented by jagoll last updated on 08/Jan/20

$$sirminimum{(3x+4)}^2=0$$$$whyminimum{(3x+4)}^2\times \sin \left(2x\right)=-\mathrm{\infty }$$

Commented by MJS last updated on 08/Jan/20

$$...\mathrm{similar}\mathrm{for}$$$$f\left(\frac{4n+1}{4}\pi \right)=\frac{{(12\pi n+3\pi +16)}^2}{16}$$

Commented by jagoll last updated on 08/Jan/20

$$thanksyousir$$

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