# Critical points of y=2cos(x) + |x| ?

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thus the picture of the function would ber classical cosinal functin with amplitude of 2 that is stretched upwards along the 45degrees upwards line crossing points [1;1] [2;2] [3;3] etc. from the yero to the left the line will be [-1;1] [-2;2] etc.so the minima would be at minima of the cos(x) that means at positions +180 degrees = pi() and -180 degrees= -pi() , both with value y=pi() -2 =approx 1.14

the points mentioned above are not the whole function, thez are just drawing the absolute x function. the minima are reffered to the entire function.

try visualizing the function in the MS excel

set one column values x -4*PI to +4*PI step PI/2

the other column make function abs(x)

the third make 2*cos(x)

the fourth =the second and the third

set up graph x values of the first column

y values of the fourth column

done

to state the minimum precisely you need to derive the function then you should get the y’=2*sin(x)+1 /i am not quite sure/ equal it to zero it means no growth=local minimum estimate x, the same is working for the negative x values

define your function as having two parts

y=2cos(x)+x for x>=0

and

y=2cos(x)-x for x=<0 the function is symmetrical (plot it with excel) Now lets investigate the positive side A) y'=-2sin(x)+1 B) y''=-2cos(x) The critical points are where A) and B) are equal to 0 for A) y'=0 implies -2sin(x)=1 sin(x)=1/2 x should be expressed in radians, there are pi radians in 180 degrees x=30+n*360 & x= 150+n*360 degrees Pi/6 Radians=30 degrees Where n is a Natural Number such as (1,2,3...... For B) y''=-2cos(x) y''=0 implies -2cos(x)=0 cos(x)=0 implies x= 90+n*360 and 270+n*360 degrees Similarly do the investigate the function for negative values of x

|x| = x; if x>0

= -x; if x<0 now try to solve the problem

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