Science:Math Exam Resources/Courses/MATH102/December 2011/Question 05/Solution 1

To determine the absolute max and min of the function f(x) on the interval we need to determine the critical points of the function and evaluate the function f(x) at those points, as well as the endpoints of the interval. The absolute max(min) of f(x) on the interval $[0,2\pi ]$ is given by the greatest(least) value of the function at those points.

To determine the critical points, we evaluate solve $f'(x)=0$ for $x$.

$0=f'(x)=\cos(x)+\sin(x).$

and thus when $\cos(x)\neq 0$

$\displaystyle -1={\frac {\sin(x)}{\cos(x)}}=\tan(x)$

If $\cos(x)$ is 0, then we know that $x={\frac {\pi }{2}}$ or $x={\frac {3\pi }{2}}$. At these points, $\sin(x)\neq 0$ and thus $0\neq \cos(x)+\sin(x)$. Hence we may suppose that $\cos(x)\neq 0$.

We know that $\displaystyle \tan(x)$ is negative in the second and fourth quadrants. We also know that a right triangle with side lengths $\displaystyle 1:1:{\sqrt {2}}$ has angles $\displaystyle {\frac {\pi }{4}}$, $\displaystyle {\frac {\pi }{4}}$ and $\displaystyle {\frac {\pi }{2}}$. Thus, we have that $\displaystyle x={\frac {3\pi }{4}}$ and $\displaystyle x={\frac {7\pi }{4}}$ (see the diagram below).

Thus, we plug in all these values to see that

$\displaystyle f(0)=\sin(0)-\cos(0)=-1$

$\displaystyle f({\tfrac {3\pi }{4}})=\sin({\tfrac {3\pi }{4}})-\cos({\tfrac {3\pi }{4}})={\frac {1}{\sqrt {2}}}-{\frac {-1}{\sqrt {2}}}={\frac {2}{\sqrt {2}}}={\sqrt {2}}$

$\displaystyle f({\tfrac {7\pi }{4}})=\sin({\tfrac {7\pi }{4}})-\cos({\tfrac {7\pi }{4}})={\frac {-1}{\sqrt {2}}}-{\frac {1}{\sqrt {2}}}={\frac {-2}{\sqrt {2}}}=-{\sqrt {2}}$

$\displaystyle f(2\pi )=\sin(2\pi )-\cos(2\pi )=0-1=-1$

and thus, the maximum occurs at $\displaystyle {\tfrac {3\pi }{4}}$ and its value is $\displaystyle {\sqrt {2}}$ and the minimum occurs at $\displaystyle {\tfrac {7\pi }{4}}$ and its value is $\displaystyle -{\sqrt {2}}$