What are the absolute extrema of #f(x)=2cosx+sinx in[0,pi/2]#?

1 Answer
May 1, 2017

Answer:

Absolute max is at #f(.4636) approx 2.2361#

Absolute min is at #f(pi/2)=1#

Explanation:

#f(x)=2cosx+sinx#

Find #f'(x)# by differentiating #f(x)#
#f'(x)=-2sinx+cosx#

Find any relative extrema by setting #f'(x)# equal to #0#:
#0=-2sinx+cosx#

#2sinx=cosx#

On the given interval, the only place that #f'(x)# changes sign (using a calculator) is at

#x=.4636476#

Now test the #x# values by plugging them into #f(x)#, and don't forget to include the bounds #x=0# and #x=pi/2#

#f(0) = 2#

#color(blue)(f(.4636) approx 2.236068)#

#color(red)(f(pi/2) = 1)#

Therefore, the absolute maximum of #f(x)# for #x in [0,pi/2]# is at #color(blue)(f(.4636) approx 2.2361)#, and the absolute minimum of #f(x)# on the interval is at #color(red)(f(pi/2)=1)#