- #1
Kamataat
- 137
- 0
Find the interval in which the function y=x+\sin x\cos x is increasing. So, first I differentiated to get y'=1+\cos 2x. Then I set y' equal to zero:
1+\cos 2x=0
\cos 2x=-1
2x=\pm \arccos m+2n\pi, where n\in\mathbb{Z}
2x=\pm \arccos(-1)+2n\pi
2x=\pm\pi+2n\pi
x=\pm\frac{\pi}{2}+n\pi
So, since y'=0 is true only at certain points (because n\in\mathbb{Z}), we know that the function is strictly increasing or decreasing. To find out which, we do this:
x_1=-50 : y(x_1)=y_1=-49.75
x_2=30 : y(x_2)=y_2=29.85
Thus the function is strictly increasing because in the case of x_1 < x_2 we have y_1 < y_2.
So the function is increasing on the open interval X^{\uparrow}=]-\infty;\infty[.
Is this correct?
- Kamataat
1+\cos 2x=0
\cos 2x=-1
2x=\pm \arccos m+2n\pi, where n\in\mathbb{Z}
2x=\pm \arccos(-1)+2n\pi
2x=\pm\pi+2n\pi
x=\pm\frac{\pi}{2}+n\pi
So, since y'=0 is true only at certain points (because n\in\mathbb{Z}), we know that the function is strictly increasing or decreasing. To find out which, we do this:
x_1=-50 : y(x_1)=y_1=-49.75
x_2=30 : y(x_2)=y_2=29.85
Thus the function is strictly increasing because in the case of x_1 < x_2 we have y_1 < y_2.
So the function is increasing on the open interval X^{\uparrow}=]-\infty;\infty[.
Is this correct?
- Kamataat
.
