Question

Find the absolute extrema of the function f(x)=2sinx-cos(2x) on the closed interval [0,3π/2]

plz help me​

Answer 1

2cosx +sin2x=0

substitute x= 0,π/2

-1

3

-2

means at this x =π/2 it has absolute Maxima

ummmmma

Answer 2

The absolute minima of the function is -1.5 and absolute maxima of the function is 3.

• Given f(x) = 2sinx-cos(2x). Taking the first derivative of f(x) we get $f^{'}(x) = 2cosx+2sin2x$
• Now to find the absolute extrema we have to find first the local extrema. Equating this to 0 we get , $2cosx+2sin2x = 0$
• We get the roots from it as $\frac{\pi}{2} , \frac{3\pi }{2} , \frac{7\pi }{6}$ .
• Also the lower and upper value of the given interval are 0 and $\frac{3\pi }{2}$ respectively.
• Now f(0) = -1 , f($\frac{3\pi }{2}$) = -1 , f($\frac{\pi }{2}$) = 3 and f($\frac{7\pi }{6}$) = -1.5
• So we get the absolute minima of the function as -1.5 and absolute maxima of the function as 3.