Absolute maximum and minimum?

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1 Answer
Mar 11, 2018

#(-pi/4,-1)# is the minimum.

#(pi/8,sqrt2/2)# is the maximum.

Explanation:

Maximum and minimum values of a curve may be one of the critical points.

Critical points are points where the instantaneous rate of change is zero or undefined.

We have:

#f(x)=sin(2x)# Let's find its derivative.

#=>f'(x)=d/dx(sin(2x))#

Recall that:

#d/dx(sinx)=cosx#

#d/dx(g(h(x)))=g'(h(x))*h'(x)#

Therefore:

#f'(x)=cos(2x)*d/dx(2x)#

Now recall the power rule:

#d/dx(x^n)=nx^(n-1)# where #n# is a constant.

We now have:

#f'(x)=cos(2x)*2#

#=>f'(x)=2cos(2x)#

Hmm... It is clear that all points on our function is differentiable.

We set #f'(x)# to #0#.

#=>0=2cos(2x)# Let's solve this equation.

#=>0=cos(2x)#

#=>Arccos(0)=2x#

#=>pi/2=2x#

#=>pi/4=x=-pi/4# If #x# is between #pi# and #-pi#

We see that #pi/4>pi/8# and #-pi/4> -pi/2#

Therefore, our only extrema that is within the bound is #(-pi/4,f(-pi/4))#

Let's find #f(-pi/4)#

#=>f(-pi/4)=sin(2*(-pi/4))#

#=>f(-pi/4)=sin(-pi/2)#

#=>f(-pi/4)=-1#

Let's try out our other extrema.

#=>f(pi/4)=sin(2*(pi/4))#

#=>f(pi/4)=sin(pi/2)#

#=>f(pi/4)=1#

Therefore, our minimum is #(-pi/4,-1)#

Since our other extrema isn't within bound, we try the boundary points.

#f(-pi/2)=sin(2*-pi/2)#

#f(-pi/2)=sin(-pi)#

#f(-pi/2)=0#

Clearly, this isn't the maximum.

#f(pi/8)=sin(2*pi/8)#

#f(pi/8)=sin(pi/4)#

#f(pi/8)=sqrt2/2#

This is our maximum point.

In conclusion:

#(-pi/4,-1)# is the minimum.

#(pi/8,sqrt2/2)# is the maximum.