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Sketch a graph of a function p(x) that satisfies the following conditions: a. p(x) has a removable discontinuity at x = -1, and is continuous everywhere else in x ∈ (-[infinity], 1) ∪ (1, [infinity]). b. p(-1) = 3 c. p(x) has an absolute maximum of 3. d. lim(x→0-) p(x) = 0 e. p(x) has a relative minimum at x = -3. f. lim(x→[infinity]) p(x) = 0

A) [No options provided]
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User Sowdri
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1 Answer

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Final answer:

To sketch the graph, address each condition one by one: a. Removable discontinuity at x = -1, b. p(-1) = 3, c. Absolute maximum of 3, d. lim(x→0-) p(x) = 0, e. Relative minimum at x = -3, f. lim(x→[infinity]) p(x) = 0.

Step-by-step explanation:

To sketch a graph of a function p(x) satisfying the given conditions, we can start by addressing each condition one by one.

a. A removable discontinuity at x = -1 means that there is a hole in the graph of the function at x = -1. We can represent this by drawing an open circle at x = -1 on the graph.

b. Since p(-1) = 3, we can plot the point (-1, 3) on the graph.

c. To have an absolute maximum of 3, the graph of p(x) should have a peak at some point. We can add a local maximum at any point we choose, as long as its y-coordinate is 3.

d. To have lim(x→0-) p(x) = 0, we can ensure that the graph approaches the x-axis from the left side as x approaches 0.

e. To have a relative minimum at x = -3, we can add a local minimum at that point.

f. To have lim(x→[infinity]) p(x) = 0, the graph needs to approach the x-axis as x goes to positive infinity.

By incorporating all these conditions, we can sketch the final graph of p(x).

User Mkrause
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