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How do you find the intervals in which the function #f(x) = sin^4x + cos^4x# is increasing and decreasing in #[0, pi/2]#?

1 Answer
Jun 29, 2017

Answer:

Please see below.

Explanation:

Investigate the sign of #f'(x)# on the interval #[0,pi/2]#

#f'(x) = 4sin^3x cosx -4cos^3xsinx#

#f'# is defined for all #x# and we need to find its zeros.

#f'(x) = 4sinxcosx(sin^2x-cos^2x) = 0#

A product is zero if and only if one of its factors is zero.

Obviously, #4# cannot be #0#

#sinx = 0# at #x=0# #" "# (Remember we have a restricted domain.)

#cosx = 0# at #x=pi/2#

#sin^2x-cos^2x = 0# at #x = pi/4# #" "# (See note below.)

In #[0,pi/4]# test #f'(pi/6)# to see that #f'# is negative, so #f# is decreasing.

In #[pi/4,pi/2]# test #f'(pi/3)# to see that #f'# is positive, so #f# is increasing.

Note

Here are three ways to solve #sin^2x-cos^2x = 0#

Use the double angle formula for cosine

#sin^2x-cos^2x = -(cos^2x-sin^2x) = -cos(2x) = 0# at #2x = pi/2+2pik# and at #x=(3pi)/2+2pik#. The only solution with #x# in #[0,pi/2]# is #x=pi/4#

Factor first

#(sinx+cosx)(sinx-cosx) = 0#

#sinx+cosx = 0# has no solutions in the first quadrant.

#sinx-cosx=0# iff #sinx = cosx#, so #x = pi/4#.

Divide first

#sin^2x = cos^2x# iff #sin^2x/(cos^2x) = 1# which requires #tanx = +-1#. In the first quadrant the only solution is #x = pi/4#.