Question #dc02f

1 Answer
Jun 4, 2017

Max:-0.464+(2k)pi
Min:-0.464+(2k+1)pi

Explanation:

Find the derivative:
2cosx-sinx
d/dx[2cosx-sinx]=2(-sinx)-cosx
=-2sinx-cosx

The maximum and minimum is achieved when f'(x) =0

0=-2sinx-cosx
cosx=-2sinx
-1/2=(sinx)/(cosx)
-1/2=tanx
tan^(-1)(-1/2)=x
x=-0.464...

A more general solution is:
x=-0.464+kpi

Where k is an integer.

Since tanx has a period of pi.

However it does differentiate the values for max and min.

Let us try when x=-0.464 that is when k=0 in the original function. This would achieve the max value since -sin(-0.464)=+sin(0.464) and cosx will be positive.
When x=-0.464+2pi it would achieve the same thing.

However when x=-0.464+pi it would make the entire function would negative

Thus:
Max:-0.464+(2k)pi
Min:-0.464+(2k+1)pi