Need helping solving for CI
1 answer:
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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:
On a unit circle, the point that corresponds to an angle of is at position
.
The point that corresponds to an angle of is at position
.
Step-by-step explanation:
On a cartesian plane, a unit circle is
- a circle of radius
,
- centered at the origin
.
The circle crosses the x- and y-axis at four points:
,
,
, and
.
Join a point on the circle with the origin using a segment. The "angle" here likely refers to the counter-clockwise angle between the positive x-axis and that segment.
When the angle is equal to , the segment overlaps with the positive x-axis. The point is on both the circle and the positive x-axis. Its coordinates would be
.
To locate the point with a angle, rotate the
segment counter-clockwise by
. The segment would land on the positive y-axis. In other words, the
-point would be at the intersection of the positive y-axis and the circle. Its coordinates would be
.
