## Precalculus (6th Edition) Blitzer

a. $98.52\ ^\circ F$ b. $24\ hours$ c. 5pm. $98.9\ ^\circ F$ d. 5am. $98.3\ ^\circ F$ e. see graph. a. From the given function, with $x=0$, we have $y(0)=98.6+0.3\ sin(\frac{\pi}{12}(0)-\frac{11\pi}{12})\approx98.6+0.3(-0.2588)\approx98.52\ ^\circ F$ b. The period of the cycle can be found as $p=\frac{2\pi}{\pi/12}=24\ hours$ c. The body temperature will reach a maximum when the sine function is $1$ which leads to $\frac{\pi}{12}(x)-\frac{11\pi}{12}=2k\pi+\frac{\pi}{2}$ or $x=24k+17$ where $k$ is an integer. As $x\in[0,24]$, we have $x=17\ hours$ after midnight or 5pm. The highest temperature is $y(17)=98.6+0.3=98.9\ ^\circ F$ d. The body temperature will reach a minimum when the sine function is $-1$ which leads to $\frac{\pi}{12}(x)-\frac{11\pi}{12}=2k\pi-\frac{\pi}{2}$ or $x=24k+5$. As $x\in[0,24]$, we have $x=5\ hours$ after midnight or 5am. The lowest temperature is $y(5)=98.6-0.3=98.3\ ^\circ F$ e. Based on the above results, we can graph the function for one period in $[0,24]$ as shown in the figure.