# Maximum and Minimum Values of a Trig Function

• Dec 6th 2011, 08:49 PM
newslang
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.

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

I know that the derivative is
$\displaystyle 2tanxsecx+sec^2x$
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
• Dec 6th 2011, 08:58 PM
Prove It
Re: Maximum and Minimum Values of a Trig Function
Quote:

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.

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

I know that the derivative is
$\displaystyle 2tanxsecx+sec^2x$
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 \displaystyle \begin{align*} f(-\pi) \end{align*} and \displaystyle \displaystyle \begin{align*} f(2\pi)\end{align*} as well.

\displaystyle \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*}