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

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! 
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2 Answers

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.
 
 
 
 
 

Comments

So my question is, in order to figure out the sign, we should plug a value into derivative. What x value did you plug on (-π/2,0)?
For any interval, you can choose any value that is within the interval, in order to determine the sign of an expression--when you have identified all of the zeros of the function being analyzed.
 
The midpoint of the interval can be used, e.g. -pi/4 for (-pi/2,0).
 
 
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.