+130511 y = sinx + cosx
y' = cosx - sinx
Setting to 0, we have
cosx - sin x = 0 add sin x to both sides
cosx = sinx and this is true at pi/4 and (5/4)pi on [0, 2pi]
And taking the second derivative, we have
y" = -sin x - cosx
So....evaluating this at pi/4, we have
-sin(pi/4) - cos(pi/4) = -(1/√2√2) - (1/√2) = -2/√2 ....so the curve is concave down at tthis point
And evaluating the second derivative at (5/4)pi, we have
-sin(5/4 pi) - cos(5/4pi) = (1/√2√2) - (-1/√2) = 2/√2 ....so the curve is concave up at this point
So....on [0, 2pi], the graph would be increasing on [0, pi/4) ....and it would be decreasing on (pi/4, 5/4 pi)..... and then it would be increasing again at (5/4 pi, 2pi]......and it would repeat this behavior on every interval [n(2pi), (n+1)(2pi)]
Here's the graph......https://www.desmos.com/calculator/twbniyyrfu
