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

what is maximum value of y= (3x+4)^2 ×sin (2x)

$${what}\:{is}\:{maximum}\: \\ $$$${value}\:{of}\:{y}=\:\left(\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} ×\mathrm{sin}\:\left(\mathrm{2}{x}\right) \\ $$

Commented by MJS last updated on 08/Jan/20

sin 2x =−1 ⇒ x=nπ−(π/4)=((4n−1)/4)π sin 2x =+1 ⇒ x=nπ+(π/4)=((4n+1)/4)π n∈Z f(x)=(3x+4)^2 sin 2x f(((4n−1)/4)π)=−(((12πn−3π+16)^2 )/(16)) n=±10^3 ⇒ f(x)≈−8.9×10^7 n=±10^6 ⇒ f(x)≈−8.9×10^(13) n=±10^9 ⇒ f(x)≈−8.9×10^(19) ...

$$\mathrm{sin}\:\mathrm{2}{x}\:=−\mathrm{1}\:\Rightarrow\:{x}={n}\pi−\frac{\pi}{\mathrm{4}}=\frac{\mathrm{4}{n}−\mathrm{1}}{\mathrm{4}}\pi \\ $$$$\mathrm{sin}\:\mathrm{2}{x}\:=+\mathrm{1}\:\Rightarrow\:{x}={n}\pi+\frac{\pi}{\mathrm{4}}=\frac{\mathrm{4}{n}+\mathrm{1}}{\mathrm{4}}\pi \\ $$$${n}\in\mathbb{Z} \\ $$$$ \\ $$$${f}\left({x}\right)=\left(\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}{x} \\ $$$${f}\left(\frac{\mathrm{4}{n}−\mathrm{1}}{\mathrm{4}}\pi\right)=−\frac{\left(\mathrm{12}\pi{n}−\mathrm{3}\pi+\mathrm{16}\right)^{\mathrm{2}} }{\mathrm{16}} \\ $$$${n}=\pm\mathrm{10}^{\mathrm{3}} \:\Rightarrow\:{f}\left({x}\right)\approx−\mathrm{8}.\mathrm{9}×\mathrm{10}^{\mathrm{7}} \\ $$$${n}=\pm\mathrm{10}^{\mathrm{6}} \:\Rightarrow\:{f}\left({x}\right)\approx−\mathrm{8}.\mathrm{9}×\mathrm{10}^{\mathrm{13}} \\ $$$${n}=\pm\mathrm{10}^{\mathrm{9}} \:\Rightarrow\:{f}\left({x}\right)\approx−\mathrm{8}.\mathrm{9}×\mathrm{10}^{\mathrm{19}} \\ $$$$... \\ $$

Commented by jagoll last updated on 08/Jan/20

please

$${please}\: \\ $$

Commented by MJS last updated on 08/Jan/20

no maximum because lim_(x→±∞) (3x+4)^2 =+∞

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

Commented by jagoll last updated on 08/Jan/20

if minimum sir?

$${if}\:{minimum}\:{sir}? \\ $$

Commented by MJS last updated on 08/Jan/20

also no minimum −1≤sin 2x ≤1 −∞<(3x+4)^2 sin 2x<+∞

$$\mathrm{also}\:\mathrm{no}\:\mathrm{minimum} \\ $$$$−\mathrm{1}\leqslant\mathrm{sin}\:\mathrm{2}{x}\:\leqslant\mathrm{1} \\ $$$$−\infty<\left(\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}{x}<+\infty \\ $$

Commented by jagoll last updated on 08/Jan/20

thanks you sir

$${thanks}\:{you}\:{sir} \\ $$

Commented by jagoll last updated on 08/Jan/20

sir minimum (3x+4)^2 = 0 why minimum (3x+4)^2 ×sin (2x) = −∞

$${sir}\:{minimum}\:\left(\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} \:=\:\mathrm{0}\: \\ $$$${why}\:{minimum}\:\left(\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} ×\mathrm{sin}\:\left(\mathrm{2}{x}\right)\:=\:−\infty \\ $$

Commented by MJS last updated on 08/Jan/20

...similar for f(((4n+1)/4)π)=(((12πn+3π+16)^2 )/(16))

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

Commented by jagoll last updated on 08/Jan/20

thanks you sir

$${thanks}\:{you}\:{sir} \\ $$

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