Question 1067370
<pre><b>
If f(x)=sinx+cos2x,determine which sub-intervals of (0,2pi)
f(x) is increasing on and which intervals f(x) is decreasing.

f(x) = sin(x) + cos(2x)
f'(x) = cos(x) - 2sin(2x)

We set f'(x)=0 to see if there are any critical
values as far as increasing or decreasing:

cos(x) - 2sin(2x) = 0

cos(x) -2[2sin(x)cos(x)] = 0

cos(x) -4sin(x)cos(x) = 0

cos(x)[1 - 4sin(x)] = 0

cos(x) = 0;   1 - 4sin(x) = 0
x = {{{matrix(1,3,pi/2,",",3pi/2)}}}; -4sin(x) = -1
                                        sin(x) = 1/4
                                             x = .2527, 2.8889

Substituting test values on each interval, we find:

It's increasing on {{{(matrix(1,3,0,",",0.2527))}}}
It's decreasing on {{{(matrix(1,3,0.2527,",",pi/2))}}}
It's increasing on  {{{(matrix(1,3,pi/2,",",2.8889))}}}
It's decreasing on {{{(matrix(1,3,2.8889,",",3pi/2))}}}
It's increasing on {{{(matrix(1,3,3pi/2,",",2pi))}}}

{{{drawing(400,3200/13,-.1,6.4,-2,2,

green(circle(.2526817,1.125,.1),circle(pi/2,0,.1), circle(2.8889146,1.125,.1),
circle(4.7123897,-2,.1)


),


graph(400,3200/13,-.1,6.4,-2,2,sin(x)+cos(2x)))}}}

The 1st green circle on the graph is where x = 0.2527
The 2nd green circle on the graph is where x = {{{pi/2}}}
The 3rd green circle on the graph is where x = 2.8889
The 4th green circle on the graph is where x = {{{3pi/2}}}

Edwin</pre></b>