Find the maximum and minimum values for the function #f# defined by #f(x) = 2sinx + cos2x# in the interval #[0, pi/2]#?

Question

Paisley Johnson · Accepted Answer

Minimum:
#f(x) = {1, x=0, pi/2}#
Maximum:
#f(x) = {3/2, x=pi/6}# #f(x) = 2sinx + cos2x# #f'(x) = 2cosx - 2sin2x# Now to find the critical number, #f'(x) = 0# #2cosx - 2sin2x = 0# #2cosx-2(2sinxcosx) = 0# #2cosx - 4sinxcosx = 0# #2cosx(1-2sinx) = 0# #2cosx=0# #x = pi/2# and #1-2sinx=0# #x=pi/6# Therefore, When #x=0#, #f(x) = 2.0+1 = 1# and when #x=pi/2, f(x) = 2.1-1=1# And when #x=pi/6, f(x) = 2.1/2+1/2 = 3/2# Hence, Minimum: #f(x) = {1, x=0, pi/2}# Maximum: #f(x) = {3/2, x=pi/6}#

Emilia Aguilar · Answer

undefined To find the maximum and minimum values of the function $ f(x) = 2\sin(x) + \cos(2x) $ in the interval $[0, \frac{\pi}{2}]$, we need to first find the critical points of the function within this interval.

The critical points occur where the derivative of the function is either zero or undefined.

First, we find the derivative of $ f(x) $:
$$ f'(x) = 2\cos(x) - 2\sin(2x) $$

Next, we find the critical points by setting the derivative equal to zero and solving for $ x $:
$$ 2\cos(x) - 2\sin(2x) = 0 $$

Solving this equation yields the critical points $ x = \frac{\pi}{6} $ and $ x = \frac{\pi}{2} $.

Next, we evaluate the function $ f(x) $ at the critical points and at the endpoints of the interval $[0, \frac{\pi}{2}]$ to find the maximum and minimum values.

$$ f(0) = 2\sin(0) + \cos(0) = 0 + 1 = 1 $$
$$ f\left(\frac{\pi}{6}\right) = 2\sin\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{3}\right) = 2\left(\frac{1}{2}\right) + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2} $$
$$ f\left(\frac{\pi}{2}\right) = 2\sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 2(1) + 0 = 2 $$

The maximum value of the function occurs at $ x = \frac{\pi}{2} $, where $ f(x) = 2 $.

The minimum value of the function occurs at $ x = 0 $, where $ f(x) = 1 $.

Therefore, the maximum value of the function in the interval $[0, \frac{\pi}{2}]$ is $ 2 $ and the minimum value is $ 1 $.