#f(x)= abs((x^2-12)(x^2+4))#, on the interval #-2<=x<=3#
I'll assume that the reader can calculate:
#(f(3)-f(-2))/(3-(-2)) = (39-64)/5 = -5#
Note that #x^2+4 > 0# for all real #x#, so we can write:
#f(x) = abs(x^2-12)(x^2+4)#
Also note that #x^2-12# changes sign at #x= +-sqrt12#. And #sqrt12 ~~ 3.5#. (Note: #3.5^2 = 3*4+0.25= 12.25#)
Therefore, the quantity #x^2-12# does not change sign in #[-2, 3]# and is, in fact, negative.
So #abs(x^2-12) = -(x^2-12) for all #x# in our interval.
On the interval #[-2, 3]#, we have, then:
#f(x) = -(x^2-12)(x^2+4) = -x^4+8x^2+48#
So, #f'(x) = -4x^3+16x#
We want to know how many solutions there are to
#-4x^3+16x = -5# in #[-2, 3]#
or
#4x^3-16x-5 = 0#
This means we want to find zeros of
#g(x) = 4x^3 - 16x - 5 = 4x(x^2-4)-5#
We're not going to try to solve the equation, just count roots.
Note that #g(x)# is continuous on every interval of reals. This tells us that we can apply the Intermediate Value Theorem on any closed interval we like.
#g(-2) = (-8)(0)-5 = -5#
#g(-1) = (-4)(-3)-5 = 7#
#g# is continuous on #[-2, -1]# so there is at least one zero between #-2# and #-1#.
#g(0) = -5#, so, there is a least a second zero between #-1# and #0#
#g(1)# and #g(2)# are also negative, but
#g(3) = 12(9-4)-5# is positive,
so there is at least one more zero between #2# and #3#.
There are at least #3# zeros in #[-2, 3]#, but a cubic can have at most #3# zeros total. We conclude that there are exactly #3# zeros for #g(x)# in #[-2, 3]#, each of which satisfies the conclusion of the Mean Value Theorem on #[-2, 3]#
If you have access to graphing technology, you can graph
#g(x) = 4x^3 - 16x - 5# to see that there are 3 zeros in the interval:
graph{4x^3 - 16x - 5 [-4.82, 6.28, -1.843, 3.706]}
