Lets write function in the form:
#f(x)=4-x^2, x in[-2,2]#
#f(x)=-4+x^2, x in (-oo,-2)uu(+2,+oo)#
1st derivative:
#f'(x)=-2x, x in[-2,2]#
#f'(x)=2x, x in (-oo,-2)uu(+2,+oo)#
#f'(x)=0#
#-2x=0 <=> x=0, x in [-2,2], x# is critical point
#2x=0 <=> x=0, x !in (-oo,-2)uu(+2,+oo), x# is not critical point
Sign of 1st derivative:
#AAx in (-oo,-2) f'(x)=2x<0, f# is decreasing
#AAx in (-2,0) f'(x)=-2x>0, f# is increasing
#AAx in (0,+2) f'(x)=-2x<0, f# is decreasing
#AAx in (+2,+oo) f'(x)=2x>0, f# is increasing
#f'(x)=2x# on the interval #[-2,2]# is continuous function, #f'(x)# changes sign in #x=0# hence function #f(x)# has maximum value in #x=0# and #f_max=f(0)=4#.
In points #x=-2# and #x=2# may exist discontinuities of the 1st order, so we have to check if the function #f(x)# is continuous in these points.
#f(-2)=0#
#lim_(x->-2_-)f(x)=lim_(x->-2_-)(-4+x^2)=A#
#x=-2-epsilon#
#A=lim_(epsilon->0)(-4+(-2-epsilon)^2)=lim_(epsilon->0)(-4+4+4epsilon+epsilon^2)#
#=lim_(epsilon->0)(4epsilon+epsilon^2)=0#
We prove that #lim_(x->2_-)f(x)=f(-2)# and hence function is continuous at the point #x=-2#.
#f'(x)# changes sign in #x=-2# and function has minimum value #f_min=f(-2)=0#.
#f(2)=0#
#lim_(x->2_+)f(x)=lim_(x->2_+)(-4+x^2)=B#
#x=2+epsilon#
#B=lim_(epsilon->0)(-4+(2+epsilon)^2)=lim_(epsilon->0)(-4+4+4epsilon+epsilon^2)#
#=lim_(epsilon->0)(4epsilon+epsilon^2)=0#
Again, we prove that #lim_(x->2_+)f(x)=f(2)# and hence function is continuous at the point #x=2#.
#f'(x)# changes sign in #x=2# and function has minimum value #f_min=f(2)=0#.
