Question

determine the maximum value, the minimum value, and the values of x where they occur for each function on the interval ( xin-2pi,2pi ).
a) ( y = sin x - 5 )
b) ( y=cos x + 7 )
c) ( y=sin x + 1 )
d) ( y=cos x - 3 )

Explanation:

Step1: Recall the range of basic sine and cosine functions

The range of \(y = \sin x\) is \([- 1,1]\), and the range of \(y=\cos x\) is \([-1,1]\).

Step2: Analyze the transformation for \(y = \sin x-5\)

For the function \(y=\sin x - 5\), we use the transformation \(y = f(x)+k\). Here \(f(x)=\sin x\) and \(k=-5\).
The maximum value of \(\sin x\) is \(1\). So \(y_{max}=1 - 5=-4\). When \(\sin x = 1\), \(x=\frac{\pi}{2}+2n\pi\), for \(n\in\mathbb{Z}\). In the interval \(x\in[-2\pi,2\pi]\), \(x =-\frac{3\pi}{2},\frac{\pi}{2}\).
The minimum value of \(\sin x\) is \(-1\). So \(y_{min}=-1 - 5=-6\). When \(\sin x=-1\), \(x =-\frac{\pi}{2}+2n\pi\), for \(n\in\mathbb{Z}\). In the interval \(x\in[-2\pi,2\pi]\), \(x=-\frac{\pi}{2},\frac{3\pi}{2}\).

Step3: Analyze the transformation for \(y=\cos x + 7\)

For the function \(y=\cos x+7\), \(f(x)=\cos x\) and \(k = 7\).
The maximum value of \(\cos x\) is \(1\). So \(y_{max}=1 + 7=8\). When \(\cos x = 1\), \(x = 2n\pi\), for \(n\in\mathbb{Z}\). In the interval \(x\in[-2\pi,2\pi]\), \(x=-2\pi,0,2\pi\).
The minimum value of \(\cos x\) is \(-1\). So \(y_{min}=-1 + 7=6\). When \(\cos x=-1\), \(x=(2n + 1)\pi\), for \(n\in\mathbb{Z}\). In the interval \(x\in[-2\pi,2\pi]\), \(x=-\pi,\pi\).

Step4: Analyze the transformation for \(y=\sin x+1\)

For the function \(y=\sin x + 1\), \(f(x)=\sin x\) and \(k = 1\).
The maximum value of \(\sin x\) is \(1\). So \(y_{max}=1 + 1=2\). When \(\sin x = 1\), \(x=\frac{\pi}{2}+2n\pi\), for \(n\in\mathbb{Z}\). In the interval \(x\in[-2\pi,2\pi]\), \(x=-\frac{3\pi}{2},\frac{\pi}{2}\).
The minimum value of \(\sin x\) is \(-1\). So \(y_{min}=-1+1 = 0\). When \(\sin x=-1\), \(x=-\frac{\pi}{2}+2n\pi\), for \(n\in\mathbb{Z}\). In the interval \(x\in[-2\pi,2\pi]\), \(x=-\frac{\pi}{2},\frac{3\pi}{2}\).

Step5: Analyze the transformation for \(y=\cos x-3\)

For the function \(y=\cos x-3\), \(f(x)=\cos x\) and \(k=-3\).
The maximum value of \(\cos x\) is \(1\). So \(y_{max}=1-3=-2\). When \(\cos x = 1\), \(x = 2n\pi\), for \(n\in\mathbb{Z}\). In the interval \(x\in[-2\pi,2\pi]\), \(x=-2\pi,0,2\pi\).
The minimum value of \(\cos x\) is \(-1\). So \(y_{min}=-1-3=-4\). When \(\cos x=-1\), \(x=(2n + 1)\pi\), for \(n\in\mathbb{Z}\). In the interval \(x\in[-2\pi,2\pi]\), \(x=-\pi,\pi\).

Answer:

a) Maximum value: \(-4\) at \(x =-\frac{3\pi}{2},\frac{\pi}{2}\); Minimum value: \(-6\) at \(x=-\frac{\pi}{2},\frac{3\pi}{2}\)
b) Maximum value: \(8\) at \(x=-2\pi,0,2\pi\); Minimum value: \(6\) at \(x =-\pi,\pi\)
c) Maximum value: \(2\) at \(x=-\frac{3\pi}{2},\frac{\pi}{2}\); Minimum value: \(0\) at \(x=-\frac{\pi}{2},\frac{3\pi}{2}\)
d) Maximum value: \(-2\) at \(x=-2\pi,0,2\pi\); Minimum value: \(-4\) at \(x=-\pi,\pi\)