Answer:
When we have a function g(x)
We will have a maximum at x when:
g'(x) = 0
g''(x) < 0.
Now we start with:
g(x) = 3*x^4 - 8*x^3
The first derivation is:
g'(x) = 4*3*x^3 - 3*8*x^2
g'(x) = 12*x^3 - 24*x^2
We can rewrite this as:
g'(x) = x^2*(12*x - 24)
Now we want to solve:
g'(x) = 0 = x^2*(12*x - 24)
We have one trivial solution, that is when x = 0.
The other solution will be when the term inside of the parentheses is equal to zero.
Then we need to solve:
12*x - 24 = 0
12*x = 24
x = 24/12 = 2
Then g'(x) is equal to zero for x = 0, and x = 2.
Notice that both of these points are included in the interval [ -2,2 ]
Now we need to look at the second derivative of g(x):
g''(x) = 3*12*x^2 - 2*24*x
g''(x) = 36*x^2 - 48*x
Ok, now we need to evaluate this in the two roots we found before:
if x = 0:
g''(0) = 36*0^2 - 48*0 = 0
g''(0) = 0
Then we do not have a maximum at x = 0, this is a point of inflection.
if x = 2:
g''(2) = 36*(2^2) - 48*2 = 48
then:
g''(2) > 0
This is an absolute minimum.
Now let's look only at the interval [ -2,2 ]
We know that:
g''(0) = 0
g''(2) > 0
Then at some point, we should have g''(x) < 0 in our interval.
We need to find the first point such that happens, so let's try with the lower limit of the interval but with the first derivation, if g'(x) < 0, this means that the function is decreasing from that point on, then that point will be a maximum in our interval.
x = -2
g'(-2) = (-2)^2*(12*-2 - 24) = -192
And if we look at the function:
g'(x) = x^2*(12*x - 24)
We can easily see that it is negative unitl x = 0, and then it keeps being negative until x = 2.
So in the interval [ -2,2 ], the function g'(x) is always negative or zero, this means that in the interval [ -2,2 ], the function g(x) is always decreasing or constant.
Then the absolute maximum of g(x) in the interval [ -2,2 ] will be x = -2
this means that:
g(-2) ≥ g(x) for all x ∈ [ -2,2 ]