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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

### 2 Answers by Expert Tutors

Kenneth S. | Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018Expert Help in Algebra/Trig/(Pre)calculu...
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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.

What x value did you put in xcosx for (-π/2,0) ?
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).

Michael J. | Effective High School STEM Tutor & CUNY Math Peer LeaderEffective High School STEM Tutor & CUNY ...
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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.