# Critical numbers of Trig functions

• Oct 18th 2009, 05:24 AM
Open that Hampster!
Critical numbers of Trig functions
Find the critical numbers on 0 < x < 2

$f(x) = 4x - 4tan(x)$
$f'(x) = 4 - sec^2(x)$

f'(x) is undefined at $\frac{\pi}{2}$ and $\pi + \frac{\pi}{2}$

I figured the next answer was $\frac{\pi}{3}$ and $\pi + \frac{\pi}{3}$ because that makes the derivative 0, but $n\pi + \frac{\pi}{3}$ isn't in the domain of the original function.

So how would I go about finding the 0's?
• Oct 18th 2009, 05:38 AM
skeeter
Quote:

Originally Posted by Open that Hampster!
Find the critical numbers on 0 < x < 2

$f(x) = 4x - 4tan(x)$
$f'(x) = 4 - sec^2(x)$

correction ... $\textcolor{red}{f'(x) = 4 - 4\sec^2{x}}$

yes, the original function and, therefore, its derivative are undefined at $x = \frac{\pi}{2}$

$4(1-\sec^2{x}) = 0$

$\sec^2{x} = \pm 1$

only solution in the interval [0,2] is x = 0
• Oct 18th 2009, 06:25 AM
Open that Hampster!
Ugh, I was messing the the math tags and it must have disappeared.

The upper bound is 2 pi, not 2.
• Oct 18th 2009, 06:33 AM
skeeter
Quote:

Originally Posted by Open that Hampster!
Ugh, I was messing the the math tags and it must have disappeared.

The upper bound is 2 pi, not 2.

ok, then ...

the original function and its derivative are undefined at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$

$\sec^2{x} = \pm 1$ at $x = 0$ , $x = \pi$ , and $x = 2\pi$