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.
