Find the exact location of all the relative and absolute extrema of the function. HINT [See Examples 1 and 2.] (Order your answers from smallest to largest t.) h(t) = 8t3 − 24t2 with domain [−1, +[infinity])

(-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

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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.

5)

5 - (49 ÷ 1 - 5) × 2 + 9 - 3
5 - 44 x 2 + 9 - 3
5 - 88 + 9 - 3
-83 + 9 - 3
-74 - 3
= -77

6)

(2 + 76 ÷ 1 + 9) + 4 - 3 × 9
(2 + 76 + 9) + 4 - 3 × 9
87 + 4 - 3 × 9
87 + 4 - 27
91 - 27
= 64

7)

5 × 5 - 7 + 82 ÷ (2 × 1) + 5
5 × 5 - 7 + 82 ÷ 2 + 5
25 - 7 + 82 ÷ 2 + 5
25 - 7 + 41 + 5
18 + 41 + 5
59 + 5
= 64

8)

7 × 7 + 1 + 16 ÷ 8 - (6 - 7)
7 × 7 + 1 + 16 ÷ 8 - 6 + 7
49 + 1 + 16 ÷ 8 - 6 + 7
49 + 1 + 2 - 6 + 7
50 + 2 - 6 + 7
52 - 6 + 7
46 + 7
= 53

♧