Find maxima and minima of f(x)= 5sinx + 5cosx on a interval of [0,2pi]?

1 Answer
Jacob T.
May 13, 2018

There's

  • a local maximum at #(pi/2, 5)# and
  • a local minimum at #((3pi)/2,-5)#

Explanation:

#color(darkblue)(sin (pi/4))=color(darkblue)(cos (pi/4))=color(darkblue)(1)#

#f(x)=5sinx+5cosx#
#color(white)(f(x))=5(color(darkblue)(1)*sinx+color(darkblue)(1)*cosx)#
#color(white)(f(x))=5(color(darkblue)(cos(pi/4))*sinx+color(darkblue)(sin(pi/4))*cosx)#

Apply the compound angle identity for the sine function

#sin(alpha+beta)=sin alpha*cos beta+cos alpha*sin beta#

#color(black)(f(x))=5*sin(pi/4+x)#

Let #x# be the #x-#coordinate of local extrema of this function.

#5*cos(pi/4+x)=f'(x)=0#
#pi/4+x=pi/2+k*pi# where #k# an integer.
#x=-pi/2+k*pi#

#x in {pi/2, (3pi)/2}#

  • #f(pi/2)=5*sin(pi/2)=5#,
    hence there's a local maximum at #(pi/2, 5)#
  • #f(pi/2)=5*sin((3pi)/2)=-5#,
    hence there's a local minimum at #(pi/2, -5)#
Related questions
Impact of this question
3053 views around the world
You can reuse this answer
Creative Commons License