the function h is defined as h(θ) = tan θ. describe the behavior of the function h on the interval π < θ <…

Answer:

A. $h$ is increasing and concave up

Explanation:

Step1: Recall derivative of tangent

The derivative of $y = \tan\theta$ is $y'=\sec^{2}\theta$. On the interval $\pi<\theta<\frac{3\pi}{2}$, $\sec^{2}\theta> 0$. Since $y'>0$, the function $h(\theta)=\tan\theta$ is increasing.

Step2: Recall second - derivative of tangent

The second - derivative of $y = \tan\theta$: $y'=\sec^{2}\theta$, and using the chain - rule, $y'' = 2\sec\theta\cdot\sec\theta\tan\theta=2\sec^{2}\theta\tan\theta$. On the interval $\pi<\theta<\frac{3\pi}{2}$, $\sec\theta<0$ and $\tan\theta>0$, so $y''>0$. When $y''>0$, the function is concave up.