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Chapter 3: Problem 15

Find the points of inflection and discuss the concavity of the graph of the function. \(f(x)=\sin \frac{x}{2}, \quad[0,4 \pi]\)

Short Answer

Expert verified
The points of inflection of the function, \(f(x)=\sin \frac{x}{2}\), on the interval \([0,4\pi]\) are at \(x=0\) and \(x=4\pi\). The function is concave down over this interval.

Step by step solution

01

Compute the Derivative

The derivative of \(f(x)\) is given by the formula \(f'(x)=\cos \frac{x}{2} \times \frac{1}{2}\). So, \(f'(x)=\frac{1}{2}\cos(\frac{x}{2})\).
02

Compute the Second Derivative

Following the rule for differentiation, the second derivative \(f''(x)\) is given by \(-\frac{1}{4}\sin(\frac{x}{2})\). It is the second derivative that will provide the information needed to find points of inflection and discuss the concavity of the graph.
03

Find the Points of Inflection

The points of inflection are the roots of the second derivative. We solve \(-\frac{1}{4}\sin(\frac{x}{2})=0\) to find the points of inflection. Since the sine function is zero at \(0\) and \(n\pi\), where \(n\) is an integer, the equation will give the solutions: \(x=0\), \(4\pi\), \(8\pi\), etc. However, according to the given domain \([0,4\pi]\), within this interval, \(x=0\) and \(x=4\pi\) are the only points of inflection.
04

Discuss the Concavity of the Graph

The concavity of the graph is determined by the sign of the second derivative. For \(f''(x)>0\), the graph is concave up, and for \(f''(x)<0\), the graph is concave down. Between these points of inflection (from Step 3), we check a test value in the second derivative equation. For \(x\) in range \(0\) to \(4\pi\), pick the test point \(2\pi\) and substitute into \(-\frac{1}{4}\sin(\frac{x}{2})\). This will give a negative result, showing that the graph is concave down within this interval.

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