`f(x) = sqrt(x) + cos(x/2), [0,2pi]` Use a graphing utility to graph the function and find the absolute extrema of the function on the given interval.

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thilina-g | College Teacher | (Level 1) Educator

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See the attached MATLAB and it shows that there is a maximum and a minimum within that range.

We can find the x  values of these extreme points by solving the derivative of the function.

>> f = sqrt(x) + cos(x/2)
 
f =
 
cos(x/2) + x^(1/2)
 
>> ezplot(f,[0, 2*pi])
>> f1=diff(f)
 
f1 =
 
1/(2*x^(1/2)) - sin(x/2)/2

It is not possible to symbolically solve this MATLAB. But we can estimate the maximu is at x = 1.7 and the minimu is at x = 5.4 radians.

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