Problem 15 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

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 1: Compute the Derivative

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

Step 2: 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.

Step 3: 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.

Step 4: 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.