How do you solve this calculus problem?

Question

Find all extrema in the closed interval (0, 2π) if y=x+sinx

Answer 1

I'd like to point out that parenthesis are generally used for open intervals, while square brackets [ ] are used to denote closed intervals (endpoint included).    In this way, half-open intervals can be expressed as well.  For example, (a,b] is a half-open interval,  a is not included in the interval, b is included.   So  [0,2π] is the proper notation for the closed interval.   

 

y = x + sin(x)

 

Taking the derivative with respect to x:

y' = 1 + cos(x)

Setting y' = 0:

y' = 0 = 1 + cos(x)

          -1 = cos(x)

           cos(x) is -1 when x=π radians  (=180^o)

 

Now we have three values of x to check.  At this point, you may be asking: "Why are there THREE values of x to check?"

 

When you are checking for extrema on a closed interval, you must also compare f(x) at the endpoints of the interval.   So the three values of x we need to check y at are  0, π, and 2π:

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x = 0  —> y = 0 + sin(0) = 0

x = π  —>  y = π + sin(π) = π + 0 = π 

x = 2π —>  y = 2π + sin(2π) = 2π 

 

 

 

So the extrema are:  

y=0 when x=0   (the minimum on the closed interval)

and

y=2π when x=2π     (the maximum on the closed interval)

 

Notice that the value of y where y'=0 is not one of the extreme values.  Yes, that can happen.

 

 

  

 

 

 

 

 

Answer 2

Find all extrema in the closed interval (0, 2π) if y=x+sin(x)

f(x)=x+sin(x)

f'(x)=1+cos(x)

if f'(x)=0:

0=1+cos(x)

-1=cos(x)

cos(x)=-1, when x=π

f(π)=π+sin(π)=π+0=π

~3.14