On Babylonian tablet YBC 4652, a problem is given that translates to this equation:
2 answers:
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Answer:
Step-by-step explanation:
LHS = Sin² ∅ + Sin² ∅*tan² ∅
= Sin² ∅ [1 + tan² ∅ ] {Sec² ∅ - tan² ∅ = 1}
= Sin² ∅ *Sec² ∅
= Sin² ∅ *
= Sin² ∅ /Cos² ∅
= tan² ∅ = RHS
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.
