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]}
