Find the interval in which the function [itex]y=x+\sin x\cos x[/itex] is increasing. So, first I differentiated to get [itex]y'=1+\cos 2x[/itex]. Then I set [itex]y'[/itex] equal to zero:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]1+\cos 2x=0[/tex]

[tex]\cos 2x=-1[/tex]

[tex]2x=\pm \arccos m+2n\pi[/tex], where [tex]n\in\mathbb{Z}[/tex]

[tex]2x=\pm \arccos(-1)+2n\pi[/tex]

[tex]2x=\pm\pi+2n\pi[/tex]

[tex]x=\pm\frac{\pi}{2}+n\pi[/tex]

So, since [itex]y'=0[/itex] is true only at certain points (because [itex]n\in\mathbb{Z}[/itex]), we know that the function is strictly increasing or decreasing. To find out which, we do this:

[tex]x_1=-50 : y(x_1)=y_1=-49.75[/tex]

[tex]x_2=30 : y(x_2)=y_2=29.85[/tex]

Thus the function is strictly increasing because in the case of [itex]x_1 < x_2[/itex] we have [itex]y_1 < y_2[/itex].

So the function is increasing on the open interval [itex]X^{\uparrow}=]-\infty;\infty[[/itex].

Is this correct?

- Kamataat

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# Homework Help: Is this correct?

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