Answer:
Absolute maximum: 4
f(-3) = -2
Absolute minimum: Does not exist
Range: (-\infty, 4](−∞,4]
Domain: [-4, \infty)[−4,∞)
Is it a function? Yes
Relative Maximum(s): 2
Increasing Interval(s): (-2, 1)(−2,1) ... interval notation
Decreasing Interval(s): (-\infty, -2) \ \cup \ (1, \infty)(−∞,−2) ∪ (1,∞)
Relative Minimum(s): -4
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Explanations:
The absolute maximum occurs at the highest point. Specifically, it's the largest y output possible. In this case, it's y = 4.
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To determine the value of f(-3), we draw a vertical line through -3 on the x axis. Mark where this vertical line crosses the curve. Let's say its point P. From point P, draw a horizontal line until you reach the y axis. You should arrive at y = -2. Therefore, f(-3) = -2.
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The absolute min is similar to the absolute max, but now we're looking at the lowest y output possible. No such y value exists because the curve goes on forever downward.
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The range is the set of all possible y outputs. The range in compound inequality notation is -\infty < y \le 4−∞<y≤4 indicating y can be anything between negative infinity and 4. We can include 4. The range in interval notation is (-\infty, 4](−∞,4] . Note the use of the square bracket so that we include the 4.
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The domain is the set of x inputs possible. The smallest such input allowed is x = -4. There is no largest input because the graph goes on forever to the right. The domain is any x value such that -4 \le x < \infty−4≤x<∞ which condenses to the interval notation [-4, \infty)[−4,∞)
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This is a function because we cannot draw a single vertical line through more than one point on this curve; hence, this graph passes the vertical line test.
Put another way, any x input in the domain leads to exactly one and only one y output. This is a nonvisual way to prove we have a function.
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A relative maximum occurs at any peak or mountain region. It is relatively the highest point in the neighborhood/region of points. There's one such mountain peak and it's at (1,2). We can think of this as a vertex of sorts for an upside down parabola. So the relative max is y = 2 because we're only concerned with the y value.
Note: y = 4 is not a relative max because there aren't any points to the left of that endpoint. A relative extrema must have points to the left and right of it for it to be a valid neighborhood.
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Imagine this curve represents a roller coaster. As we move to the right, going uphill on this curve is an increasing section. That would be the interval from x = -2 to x = 1. So we'd say -2 < x < 1 which condenses to the interval notation (-2, 1). This is not to be confused with ordered pair (x,y) notation.
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Now we consider when we move downhill when we move to the right. This occurs on the intervals -\infty < x < -2−∞<x<−2 and also 1 < x < \infty1<x<∞ . We don't include any of the endpoints. This is because at x = -2 and x = 1, the cart is neither moving uphill nor downhill. These locations are stationary resting points so to speak.
Those two inequalities mentioned convert to the interval notations (-\infty, -2)(−∞,−2) and (1, \infty)(1,∞) in that order.
Once we determined those separated disjoint regions, we glue them together with the use of the union symbol U.
Our answer for this part would be (-\infty, -2) \ \cup \ (1, \infty)(−∞,−2) ∪ (1,∞) . Any point in this collective region will be moving downhill when moving left to right.
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This is similar to a relative maximum, but this time we're looking at the lowest valley point of a certain neighborhood. This is at (-2,-4). Therefore, the relative min is y = -4.