y = x + sin x please explain?

Question

Given the function defined by y = x + sin x for all x such that
[-pi/2, 3pi/2] for x.

(a) Find the coordinates of all maximum and minimum points on the interval. Justify your answers.

(b) Find the coordinates of all points of in
ection on the given interval. Justify your answers.

(c) Sketch the graph of the function, labeling all the points you found in (a) and (b).
i guess i can just do part c wod b easier after parts a and b

please explain
thank you!!

part b wants inflection points...

Answer 1

http://i174.photobucket.com/albums/w119/kaksi_guy/cusps.png

♦ thus y’=1+cos x = 0, hence cos x = -1,
hence taking our segment [-pi/2, 3pi/2] into account the only extreme point is
xm=pi, as -pi/2 Other possible x are x=-pi, 3pi, -3pi, and the like are out of segment!
Now is xm local min or max? To answer this we are to find y’’(x);
♣ y’’(xm) = -sin x; y’’(xm) =-sin(pi) =0;
this zero means that xm is a false extreme point! Thus no local extremes!
♣ other possible max/min points are points on ends of the segment;
y(-pi/2) =-pi/2 –1; y(3pi/2) =3pi/2 –1;
thus min y=-pi/2 –1, max y=3pi/2 –1;
♥ y’’=-sin x = 0, hence x1=0, x2=pi; Other possible x are x=-2pi, -pi, 2pi, 3pi, -3pi, and the like are out of segment!
y(0) =0, y(pi)= pi; thus (0, 0) and (pi, pi) are two inflection points!

Answer 2

y' = 1 +cos x
Set y' = 0 getting cos x = -1
Thus there is a critical point at x = pi It is a point of inflection.
The curve starts (-pi/2, -2.57) and increases in valuegoing through (0.0) and contiues to increase to (pi,pi) which is a point of inflection. The curve continues to rise to a value of 3.71 at x = 3pi/2. The concavity changes from concave down to concave up at the critical point.

Answer 3

1) to find maxima/ minima, you need to know where the derivaive =1.

y = x + sin(x), differentiate

y' = 1 + cos(x), set to zero, or

1 + cos(x) = 0, or

cos(x) = -1, which happens when x = pi

When x = pi, then y = x + sin(x) becomes
y= pi + sin(pi)
y = pi, so here is an extreme point at (pi, pi).

Answer 4

a)max and min occure when dy/dx = 0

y = x + sin(x)
dy/dx = 1 + cos(x)

dy/dx = 0 = 1+cos(x)
x = cos^-1(-1) = pi

therefore, the max occurs at x=pi

b) don't quite get what you are asking

c) try this
http://gcalc.net/

Answer 5

...*twitch*