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Chapter 3: Problem 17

Find the points of inflection and discuss the concavity of the graph of the function. \(f(x)=\sec \left(x-\frac{\pi}{2}\right), \quad(0,4 \pi)\)

Short Answer

Expert verified
The function \(f(x)=\sec \left(x-\frac{\pi}{2}\right)\) has no points of inflection and is concave upwards for all x in (0, 4蟺).

Step by step solution

01

Calculate the first derivative

In order to find the concavity and points of inflection, we first need to find the derivative \[f'(x)\] of the function \[f(x)=\sec \left(x-\frac{\pi}{2}\right)\]. To do that you need to use the chain rule. The derivative of sec(u) is sec(u)tan(u), where u is a function of x. So, \[f'(x) = \sec \left(x-\frac{\pi}{2}\right) \tan \left(x-\frac{\pi}{2}\right)\]
02

Find the second derivative

Next, find the second derivative, which is again a chain rule problem. You need to differentiate \[f'(x)\] with respect to x, applying product and chain rules.This gives: \[f''(x) = \sec \left(x-\frac{\pi}{2}\right) \tan^2 \left(x-\frac{\pi}{2}\right) + \sec^3 \left(x-\frac{\pi}{2}\right)\]
03

Find the points of inflection

Points of inflection occur where the second derivative equals zero or is undefined. Solve the equation \[f''(x) = 0\]. However, \[f''(x)\] doesn't have real roots. Then find where \[f''(x)\] is undefined. This occurs at \[ \frac{3\pi}{2}, \frac{7\pi}{2}\] in (0, 4蟺).
04

Discuss the concavity of the graph

To do this, choose test points in each interval defined by the points of inflection. For example, take x = 蟺/2, 2蟺, and 5蟺/2. Evaluate \[f''(x)\] at these test points. Positive value indicates that the graph is concave up on that interval, a negative value indicates that the graph is concave down. The function \[f''(x)\] is always positive, therefore the graph is concave upwards everywhere.

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