# Math Help - Maximum and Minimum Values of a Trig Function

1. ## Maximum and Minimum Values of a Trig Function

Hi all,

I'm struggling with the following maximum and minimum values question. The questions states: Find the absolute maximum and minimum values of the function.

$f(x)=2secx+tanx, -\pi\le x \le 2\pi$

I know that the derivative is
$2tanxsecx+sec^2x$
and I know I make that equal to zero to find critical numbers but I'm having trouble solving for x. A little help would be greatly appreciated. Thanks!

newslang

2. ## Re: Maximum and Minimum Values of a Trig Function

Originally Posted by newslang
Hi all,

I'm struggling with the following maximum and minimum values question. The questions states: Find the absolute maximum and minimum values of the function.

$f(x)=2secx+tanx, -\pi\le x \le 2\pi$

I know that the derivative is
$2tanxsecx+sec^2x$
and I know I make that equal to zero to find critical numbers but I'm having trouble solving for x. A little help would be greatly appreciated. Thanks!

newslang
First of all, you need to remember that global maxima and minima could occur at stationary points OR at endpoints. So you need to evaluate \displaystyle \begin{align*} f(-\pi) \end{align*} and \displaystyle \begin{align*} f(2\pi)\end{align*} as well.

\displaystyle \begin{align*} 2\tan{x}\sec{x} + \sec^2{x} &= 0 \\ 2\left(\frac{\sin{x}}{\cos{x}}\right)\left(\frac{1 }{\cos{x}}\right) + \frac{1}{\cos^2{x}} &= 0 \\ \frac{2\sin{x}}{\cos^2{x}} + \frac{1}{\cos^2{x}} &= 0 \\ \frac{2\sin{x} + 1}{\cos^2{x}} &= 0 \\ 2\sin{x} + 1 &= 0 \\ 2\sin{x} &= -1 \\ \sin{x} &= -\frac{1}{2} \\ x &= \left\{ \pi + \frac{\pi}{6}, 2\pi - \frac{\pi}{6} \right\} + 2\pi n \textrm{ where }n \in \mathbf{Z} \\ x &= \left\{ -\frac{5\pi}{6}, -\frac{\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\right\} \textrm{ in the interval } -\pi \leq x \leq 2\pi \end{align*}