# The maximum and minimum values of the function 1+2sinx+2cos2x in(0,pi/2) where() are closed brackets.

Rinkoo Gupta
9 years ago
f(x)=1+2sinx+2cos2x
f’(x)=2cosx-4cosxsinx=0
2cosx(1-2sinx)=0
cosx=0 or sinx=1/2
x=pi/2 or s=pi/6
pi/6 belongs to (0,pi/2) so we check this point.
f”(x)=-2sinx-4cos2x
f”(pi/6)=-2sin(pi/6)-4cos(pi/3)
=-2(1/2)-4(1/2)
=-1-2=-3 which is negative
so function is maximum at x=pi/6
so f(pi/6)=1+2sin(pi/6)+2cos2(pi/6)
=1+2(1/2)+2(sqrt3/2)2
=1+1+2(3/4)
=2+3/2=7/2
Thanks & Regards
Rinkoo Gupta
Jitender Singh IIT Delhi
9 years ago
Ans:
$f(x)=1+2sinx+2cos^{2}x$
Lets find the critical points
$f^{'}(x)=0$
$f^{'}(x)=2cosx+4cosx(-sinx)=0$
$2cosx(1-2sinx)=0$
$cosx=0, sinx = \frac{1}{2}$
$x = \frac{\pi }{6}, \frac{\pi }{2}$
To find local minima & maxima, we will find second derivative at critical points
$f^{'}(x) = 2cosx-2sin2x$
$f^{''}(x) = -2sinx-4cos2x$
$f^{''}(\frac{\pi }{6}) = -2sin(\frac{\pi }{6})-4cos2(\frac{\pi }{6}) = -3$
$f^{''}(\frac{\pi }{2}) = -2sin(\frac{\pi }{2})-4cos2(\frac{\pi }{2}) = 2$
$x = \frac{\pi }{2}$
is the minima
$x = \frac{\pi }{6}$
is the maxima
Maximum value of function:
$f(\frac{\pi }{6}) = \frac{7}{2}$
Minimumvalue of function:
$f(\frac{\pi }{2}) = 3$
Thanks & Regards
Jitender Singh
IIT Delhi