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Chapter 3: Problem 18

Find the points of inflection and discuss the concavity of the graph of the function. \(f(x)=x+2 \cos x, \quad[0,2 \pi]\)

Short Answer

Expert verified
The points of inflection of the function \(f(x) = x + 2\cos x\) are at \(x = \pi/6, 5\pi/6\). The function is concave up for \(\pi/6 < x < 5\pi/6\) and concave down for the intervals \(0 < x < \pi/6\) and \(5\pi/6 < x < 2\pi\).

Step by step solution

01

Calculate the derivative

Compute the derivative of \(f(x) = x + 2\cos x\). Using the rules of differentiation, we obtain the first derivative of the function: \(f'(x) = 1 - 2\sin x\). This represents the slope of the line tangent to the function. The sign of the first derivative gives indications about the increasing or decreasing nature of the function.
02

Calculate the second derivative

Next, compute the second derivative of the function \(f(x)\). Given that \(f'(x) = 1 -2\sin x\), then the second derivative would be \(f''(x) = -2\cos x\). This indicates the rate of change of the slope of the function. The sign of the second derivative gives us information about the concavity of the function.
03

Find the critical points

A critical point occurs where the derivative is either zero or undefined. Here, for the given function, set \(f'(x) = 0\) to find the critical points. \(1 - 2\sin x = 0 \rightarrow \sin x = 0.5\). So the critical points are \(x = \pi/6, 5\pi/6\).
04

Apply the second derivative test

Our critical points are \(x = \pi/6, 5\pi/6\). Now evaluate these points in \(f''(x)\). If \(f''(x) > 0\), we have a local minimum and the function is concave up. If \(f''(x) < 0\), we get a local maximum and the function is concave down. In this case, we get \(f''(\pi/6) = -2\cos (\pi/6) < 0\) (concave down) and \(f''(5\pi/6) = -2\cos(5\pi/6) > 0\) (concave up). This means that we have a point of inflection at \(x = \pi/6, 5\pi/6\).
05

Discuss the concavity of the function

For \(0 < x < \pi/6\) and \(5\pi/6 < x < 2\pi\), \(f''(x) < 0\), so the function is concave down. Between \(\pi/6 < x < 5\pi/6\), \(f''(x) > 0\), so the function is concave up. Hence, the function changes its concavity at the points \(x = \pi/6, 5\pi/6\).

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