#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.)