Find intervals for which f is increasing and decreasing, local max and local min, concavity and inflection points? f(x)=sinx + cosx, [0,2pie]

1 Answer
Mar 22, 2018

We start by finding the first derivative.

#f'(x) = cosx - sinx#

Since this is defined on all real values of #x#, there will be no vertical tangents. However, there will be horizontal tangents, when #f'(x) =0#. These will be our critical points.

#0 = cosx- sinx#

#sinx =cosx#

The only time this happens in the given interval is at #x = pi/4# and #x= (5pi)/4#. At #x = pi/2#, we see that the derivative equals #f'(pi/2) = cos(pi/2) - sin(pi/2) = -1#, therefore the function is decreasing on the interval #(pi/4, (5pi)/4)#, and increasing on the intervals #[0, pi/4)# and #((5pi)/4, 2pi]#.

Local maximums occur when a function goes from increasing to decreasing, so we will have a local maximum at #x= pi/4#. The local minimum is when a function goes from decreasing to increasing, so at #x =(5pi)/4#.

We need to find the second derivative to determine concavity.

#f''(x) = -sinx - cosx#

Points of inflection occur when #f''(x) = 0#.

#cosx = -sinx#

This will occur at #x = (3pi)/4# and #(7pi)/4#. We always need to check on both sides of the inflection point to make sure we go from positive to negative or negative to positive. After this we can determine the intervals of concavity. Notice that at #x = pi#, the second derivative has value #f''(pi) = -sinpi - cospi = 1#, so we're concave up on the interval #((3pi)/4, (7pi)/4)#. We're concave down on the other intervals, namely where #f''(x) < 0#:

#[0, (3pi)/4)# and #((7pi)/4, 2pi]# are going to be concave down.

Here is the final graph of #f(x)#.

graph{sin x+cosx [-10, 10, -5, 5]}

Hopefully this helps!