#f# is the composition of the absolute value and a polynomial, so #f# is continuous at every real number. In particular, #f# is continuous on #[-2,3]#. So #f# satisfies the first hypothesis on this interval..
In general #absu# is differentiable expect where #u# changes signs.
#(x^2 -12)# changes signs at #x= +- sqrt2#.
#(x^2+4)# deos not changes signs on the real numbers.
So #f# is non-differentiable ony at #+- sqrt12# .
Since #3 < sqrt12 < 4#, neither of #+- sqrt12# is in #(-2,3)#
So #f# is differentiable on #(-2,3)# And #f# satisfies the second hypothesis on this interval..
The Mean Value Theorem now assures us that there is a #c# in #(-2,3)# with #f'(c) = (f(3)-f(-2))/(3-(-2))#
Actually finding the #c# is tedious.
On the interval #[-2,3]#, the values of #(x^2-12)(x^2+4)# are negative, so
#f(x) = -(x^2-12)(x^2+4) = -x^4+8x^2+48#
And #f'(x) = -4x^3+16x#
#f(3) = 39# and #f(-2) = 64#, so we need to solve
#-4x^3 +16x = (39-64)/(3-(-2)#
#-4x^3+16x = -7#
#4x^3-16x-7=0#
Now use whatever techniques you have available to solve this third degree equation. (There are two solutions in #(-2,3)# and another outside the interval.)
