Answer:
- (-1, -32) absolute minimum
- (0, 0) relative maximum
- (2, -32) absolute minimum
- (+∞, +∞) absolute maximum (or "no absolute maximum")
Step-by-step explanation:
There will be extremes at the ends of the domain interval, and at turning points where the first derivative is zero.
The derivative is ...
h'(t) = 24t^2 -48t = 24t(t -2)
This has zeros at t=0 and t=2, so that is where extremes will be located.
We can determine relative and absolute extrema by evaluating the function at the interval ends and at the turning points.
h(-1) = 8(-1)²(-1-3) = -32
h(0) = 8(0)(0-3) = 0
h(2) = 8(2²)(2 -3) = -32
h(∞) = 8(∞)³ = ∞
The absolute minimum is -32, found at t=-1 and at t=2. The absolute maximum is ∞, found at t→∞. The relative maximum is 0, found at t=0.
The extrema are ...
- (-1, -32) absolute minimum
- (0, 0) relative maximum
- (2, -32) absolute minimum
- (+∞, +∞) absolute maximum
_____
Normally, we would not list (∞, ∞) as being an absolute maximum, because it is not a specific value at a specific point. Rather, we might say there is no absolute maximum.
Answer:
1/12 is left
Step by step explanation:
Find the common denominator which is 12 then you would get 9/12 (Originally 3/4) and 8/12 (Originally 2/3) and then you subtract which gets you 1/12
So, this question is basically asking us "If we had an x instead of a 2, would this be true?" We can try and see what we get:

So, if we want to show this we have to change the numerator or denominator in such a way that we can cancel some common factors. Notice that 
If we replace the factored numerator with the original one, we get:

Since we have an equality, this relation is proved.
what level is this question?