Maximum and Minimum Values of a Trig Function

**newslang** · December 6th 2011, 08:49 PM

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

**Prove It** · December 6th 2011, 08:58 PM

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.

Now to help you evaluate the stationary points

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

Now you need to use the second derivative test to determine if they are maxima or minima.

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