y=x+sinx...?

Question

Given the function defined by y = x + sin x for all x such that
-pi/2 is less than or equal to x which is less than or equal to 3pi/2.

(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.

Answer 1

y = x + sinx
dy/dx = 1 + cosx
set dy/dx = 0,
cosx = -1
x = pi

d^2(y)/dx^2 = -sinx
y''(pi) = -sin(pi) = 0
since the second derivative is zero for x = pi, and that the sign of y'' changes -ve to +ve, the graph should shift from concave down to concave up and thus x =pi is a pt of inflection.

a.) There are no maxima/minima critical points in the interval.
But, when we substitute the end pts of the interval, the minimum value of the function is y(-pi/2) = -pi/2 - 1, and the maximum value is y(3pi/2) = 3pi/2 - 1.

b.) x = pi, is a point of inflection.

Answer 2

We have y = x + sin(x), y' = 1 + cos(x). Since cos(x) is in [-1, 1], y' >=0 for every x in (-pi/2 3pi/2), which stands for the whole circle. So, though y' vanishes at x = pi, the function is increasing. It's sign doesn't become negative. Since (-pi/2 3pi/2), is open, y has no minimum or maximum points in this interval.

The 2nd derivative is y'' = -sin(x), which vanishes at x =0 and x = pi. The 3rd derivative is y'''= -cos(x), which is -1 at x= 0 and 1 at x = pi, both different from 0. So, we have inflections at these points. At x=0, the derivative reaches a maximum, the curve passes from concave up to concave down; at x = pi, the derivative reaches a minimum, the curve passes from concave down to to concave down.

Answer 3

lol