find-the-critical-points-of-the-function-fxcos3x2fxcos3x-2-on-the-interval-22-fr

Question

Find the critical points of the function f(x)=cos⁡(3x)−2f(x)=\cos(3x)-2 on the interval [−π2,π2][-\frac{\pi}{2},\frac{\pi}{2}]. Identify the absolute maximum and minimum values.

Accepted Answer

Critical points: x={−π3,0,π3}x=\left\lbrace-\frac{\pi}{3},0,\frac{\pi}{3}\right\rbracefmax⁡(0)=−1f_{\max}\left(0\right)=-1fmin⁡(−π3)=−3f_{\min}\left(-\frac{\pi}{3}\right)=-3 and fmin⁡(π3)=−3f_{\min}\left(\frac{\pi}{3}\right)=-3

Answer

Critical points: x={−π3,0,π3}x=\left\lbrace-\frac{\pi}{3},0,\frac{\pi}{3}\right\rbracefmax⁡(0)=3f_{\max}\left(0\right)=3fmin⁡(−π3)=0f_{\min}\left(-\frac{\pi}{3}\right)=0 and fmin⁡(π3)=0f_{\min}\left(\frac{\pi}{3}\right)=0

Answer

Critical points: x={−π3,0}x=\left\lbrace-\frac{\pi}{3},0\right\rbracefmax⁡(0)=−1f_{\max}\left(0\right)=-1fmin⁡(−π3)=−3f_{\min}\left(-\frac{\pi}{3}\right)=-3 and fmin⁡(π3)=−3f_{\min}\left(\frac{\pi}{3}\right)=-3

Answer

Critical points: x={−π3,0,π3}x=\left\lbrace-\frac{\pi}{3},0,\frac{\pi}{3}\right\rbrace fmax⁡(π3)=−1f_{\max}\left(\frac{\pi}{3}\right)=-1fmin⁡(−π3)=−3f_{\min}\left(-\frac{\pi}{3}\right)=-3

Find the critical points of the function f(x)=cos(3x)−2f(x)=\(\cos\)(3x)-2 on the interval [−π2,π2][-\(\frac{\pi}{2}\),\(\frac{\pi}{2}\)]. Identify the absolute maximum and minimum values.

Find the critical points of the function $f (x) = \cos (3 x) - 2$ on the interval $[- \frac{π}{2}, \frac{π}{2}]$ . Identify the absolute maximum and minimum values.