Question

Determine the open intervals on which the graph is concave upward or concave downward. \(y=2 x-\tan x, \quad\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

Short answer

The function \(y=2 x-\tan x\) is concave downward on all open intervals \((- \frac{\pi}{2}, \frac{\pi}{2})\) excluding the points where \(x = 0\).

Step-by-step solution

Find the first derivative

The first derivative of \(y=2x - \tan(x)\) is given by \(y'=2-\sec^{2}(x)\) where \(\sec(x)\) is the reciprocal of \(\cos(x)\).

Find the second derivative

For finding the second derivative, the derivative of \(y'\) is taken: \(y''=-2\sec^{2}(x)\tan(x)\).

Determine the intervals of concavity

The function is concave upward where the second derivative is positive, and concave downward where it's negative. It implies that the function \(y=2 x-\tan x\) is concave downward on all open intervals \((- \frac{\pi}{2}, \frac{\pi}{2})\) excluding the points where \(\sec(x) = 0\), for which \(x = 0\).