find the local max and min values of f(x)=xsinx+cosx

Question

find the local max and min values of f(x)=xsinx+cosx for -pi≤x≤pi

 

I am so stuck at how to figuring out the sign, which is important. I did take the derivative xcosx=0  then x=0 and cosx0

 

 

Please help! 

Answer 1

f' = xcosx = 0 if x={±π/2, 0} ...three critical points to locate local extrema.

This question could have asked for absolute max and absolute min ON A CLOSED INTERVAL.

 

Arranging the three critical points on a number-line that goes from -π to π, we see that there are several intervals to consider. 

 

The leftmost interval is (-π, π/2) and during that interval f' is positive so f is increasing;

on (-π/2,0) f' is negative so f is decreasing;

on (0, π/2) f' is positive so f is increasing;

on (π/2,π) f' is negative so f is decreasing.

 

These facts mean that there's a local MIN when x=0, and local MAX at ±π/2.

 

 

 

 

 

Answer 2

f'(x) = 0

 

sinx + xcosx - sinx = 0

 

xcosx = 0

 

x = 0                        cosx = 0

 

 

The interval indicates that you start on the negative x-axis, then rotate clockwise 2π on the unit circe.

 

x = -π/2                        x = π/2

 

 

Your local max and min occur at these two x values.  You can take it from here.