Question

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

Short answer

The graph of the function \(y = x + \frac{2}{\sin x}\) is concave upward on the interval (-π, 0) and concave downward on the interval (0, π).

Step-by-step solution

Find the first derivative of the function

Apply the quotient rule and the chain rule to differentiate \(y = x + \frac{2}{\sin x}\). The derivative should be \(y' = 1 - 2\csc^2 x\).

Find the second derivative

The second derivative is the derivative of the first derivative. Differentiating \(y' = 1 - 2\csc^2 x\) gives \(y'' = 4\csc^2 x \cot x\).

Determine where the second derivative is positive or negative

Set \(y''\) to be greater or less than zero. This will lead to the solutions for the intervals where the graph is concave up or down. From \(y'' = 4\csc^2 x \cot x\), we can see that \(y''\) is undefined at \( x = 0 \). Also, \(y''\) changes sign at \( x = 0 \). Therefore, \(y'' > 0\) for \(x\) in \((-π, 0)\) and \(y'' < 0\) for \(x\) in \((0, π)\).