How do you find the absolute minimum and maximum on $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ of the function $f (x) = {sin x}^{2}$ $f(x)=sinx^2$ ?

Question

6244 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-02-20T16:37:36

Hello,

Answer. Minimum is 0 and maximum is 1.

If $x$ $x$ runs on $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2 , pi/2]$ , then $x^{2}$ $x^2$ runs on $[0, \frac{π^{2}}{4}]$ $[0,pi^2/4]$ . So, you are looking for min and max of $sin$ $sin$ on the interval $[0, \frac{π^{2}}{4}]$ $[0,pi^2/4]$ .
Because $\frac{π}{2} \in [0, \frac{π^{2}}{4}]$ $pi/2 in [0,pi^2/4]$ , the maximum is $1 = sin (\frac{π}{2})$ $1 = sin(pi/2)$ .
Because $\frac{π^{2}}{4} < π$ $pi^2/4 < pi$ , sin is positive on $[0, \frac{π^{2}}{4}]$ $[0,pi^2/4]$ , so the minimum is $0 = sin (0)$ $0= sin(0)$ .

How do you find the absolute minimum and maximum on [−π2,π2][-pi/2,pi/2] of the function f(x)=sinx2f(x)=sinx^2?