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Chapter 4: Problem 87

Sketch a graph of the function. $$h(v)=\arccos \frac{v}{2}$$

Short Answer

Expert verified
The graph of the function \(h(v)=\arccos \frac{v}{2}\) starts at the point (-2,蟺), goes through (0,蟺/2) and ends at the point (2,0). The graph is undefined for values of \(v\) outside the interval [-2,2].

Step by step solution

01

Understand the function

The function is \(h(v)=\arccos \frac{v}{2}\). This is an inverse cosine function, or arccosine function. The standard cosine function has domain \([-\pi,\pi]\) and range \([-1,1]\), so the arccosine function will have domain \([-1,1]\) and range \([0,\pi]\). But because we operate with \(v/2\) instead of \(v\), our domain changes to \([-2,2]\) to accommodate that division by 2.
02

Determine key points

Key points for cosine and its inverse functions in general are at maximum, minimum and zero-crossing points. For the given function, when \(v = 2\), \(h(v)=\arccos 1 = 0\), when \(v = -2\), \(h(v)=\arccos -1 = \pi\), and when \(v = 0\), \(h(v)=\arccos 0 = \frac{\pi}{2}\).
03

Sketch the graph

Sketch the function using the key points determined in the last step. Plot the points \((-2,\pi)\), \((0,\frac{\pi}{2})\), and \((2,0)\) on the graph. The graph decreases from the point \((-2,\pi)\) to \((2,0)\). It is also important to note that for values outside the range \([-2,2]\) there is no real solution, the function is undefined.

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