The difference between the greatest and least | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given function is $f(x) = -x + \sin 2x$ on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.First,find the critical points by setting $f'(x) = 0$.$f

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(A) Given function is $f(x) = -x + \sin 2x$ on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.First,find the critical points by setting $f'(x) = 0$.$f'(x) = -1 + 2 \cos 2x = 0$$\cos 2x = \frac{1}{2}$Since $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$2x \in [-\pi, \pi]$.Thus,$2x = \pm \frac{\pi}{3}$,which gives $x = \pm \frac{\pi}{6}$.Now,evaluate $f(x)$ at the critical points and the endpoints:$f(-\frac{\pi}{2}) = -(-\frac{\pi}{2}) + \sin(-\pi) = \frac{\pi}{2} + 0 = \frac{\pi}{2} \approx 1.57$$f(-\frac{\pi}{6}) = -(-\frac{\pi}{6}) + \sin(-\frac{\pi}{3}) = \frac{\pi}{6} - \frac{\sqrt{3}}{2} \approx 0.52 - 0.866 = -0.346$$f(\frac{\pi}{6}) = -(\frac{\pi}{6}) + \sin(\frac{\pi}{3}) = -\frac{\pi}{6} + \frac{\sqrt{3}}{2} \approx -0.52 + 0.866 = 0.346$$f(\frac{\pi}{2}) = -(\frac{\pi}{2}) + \sin(\pi) = -\frac{\pi}{2} + 0 = -\frac{\pi}{2} \approx -1.57$Comparing these values,the greatest value is $M = \frac{\pi}{2}$ and the least value is $m = -\frac{\pi}{2}$.The difference is $M - m = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

The difference between the greatest and least values of the function $f(x) = -x + \sin 2x$ on $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is

Similar Questions

If $m$ and $M$ respectively denote the minimum and maximum of $f(x)=(x-1)^2+3$ for $x \in [-3, 1]$, then the ordered pair $(m, M)$ is equal to

The necessary condition for a function to have a maximum or minimum value is:

If $f(x)=a \log |x|+b x^2+x$ has extreme values at $x=-1$ and $x=2$,then the ordered pair $(a, b)=$

The perimeter of a sector is constant. If its area is to be maximum,the sectorial angle should be

If ${a^2}{x^4} + {b^2}{y^4} = {c^6}$,then the maximum value of $xy$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module