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Chapter 4: Problem 84

Sketch a graph of the function. $$g(t)=\arccos (t+2)$$

Short Answer

Expert verified
The graph of the function g(t)=\arccos(t+2) will start at the point (-3, \pi) and end at the point (-1, 0), decreasing throughout its domain, which is [-3,-1].

Step by step solution

01

Define the domain and the range

The domain of the operation g(t) = \arccos(t + 2) is simply the valid interval of t for which the function is defined by the inverse cosine operation. In this case, t + 2 should be between -1 and 1 (-1 \leq t + 2 \leq 1). By solving these inequalities, the domain of t is found to be -3 \leq t \leq -1. The range of the function is from 0 to \pi because the inverse cosine function outputs values in this interval.
02

Determine the nature and key points of the function

The function g(t) = \arccos(t + 2) is a decreasing function because the inverse cosine function decreases as its input increases. The key points are the endpoints of the interval [-3, -1], which map to [\pi, 0]. At t = -3, g(t) = \arccos(-3 + 2) = \arccos(-1) = \pi. At t = -1, g(t) = \arccos(-1 + 2) = \arccos(1) = 0.
03

Sketch the graph

Plot the key points and connect them with a smooth curve, remembering that the curve should be decreasing. The y-axis should span from 0 to \pi, and the x-axis from -3 to -1, since this is the domain of the function. Mark the point (-3,\pi) as the starting point of the graph and the point (-1,0) as the ending point.

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Most popular questions from this chapter

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