# What are the critical values, if any, of  f(x)= cotx-sinx+tanx in [0,2pi]?

Dec 27, 2015

There are no critical points at any of the $n \frac{\pi}{2}$ values; $f \left(x\right)$ is undefined at these values. However, critical points do appear at the zeroes of the derivative, which is an intricate function. From graph, critical points in given domain are approx. $x = .865 , x = 2.45 , x = 3.834 , x = 5.418$

#### Explanation:

Recall that a critical value of a function $f \left(x\right)$ occurs where the derivative of $f \left(x\right)$, denoted as either $f ' \left(x\right)$ or $\frac{\mathrm{df}}{\mathrm{dx}}$, is equal to 0 or is undefined. Thus, we must first find the derivative. First we should determine whether the function is continuous throughout its domain.

Recall that:

$\cot x = \frac{\cos x}{\sin} x$ and $\tan x = \frac{\sin x}{\cos} x$.

Since one cannot divide by zero, the original function is discontinuous in the neighborhood of those values for which $\sin \left(x\right)$ or $\cos \left(x\right) = 0$. This consists of any value which can be defined as $n \frac{\pi}{2}$, where $n$ is an integer. Within our domain we have five such values: $0 , \frac{\pi}{2} , 2 \frac{\pi}{2} = \pi , 3 \frac{\pi}{2} , 4 \frac{\pi}{2} = 2 \pi$. Thus, no matter what the derivative tells us, these cannot be critical points.

Remembering that the function is undefined (and thus non-differentiable) at those five values, we shall find the general derivative of the function. Recall that the derivative of $\sin x$ is $\cos x$. The derivatives of tan(x) and cot(x) will be quickly found below using the quotient rule.

$\frac{d}{\mathrm{dx}} \tan \left(x\right) = \frac{d}{\mathrm{dx}} \left(\sin \frac{x}{\cos} x\right) = \frac{{\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)}{\cos} ^ 2 x = \frac{1}{\cos} ^ 2 x = {\sec}^{2} x$
$\frac{d}{\mathrm{dx}} \cot \left(x\right) = \frac{d}{\mathrm{dx}} \left(\cos \frac{x}{\sin} x\right) = \frac{- {\sin}^{2} x - {\cos}^{2} x}{\sin} ^ 2 x = - 1 \frac{{\sin}^{2} x + {\cos}^{2} x}{\sin} ^ 2 x = - \frac{1}{\sin} ^ 2 x = - {\csc}^{2} x$

Thus, differentiating our original $f \left(x\right)$ yields $\frac{d}{\mathrm{dx}} \left(\cot x - \sin x + \tan x\right) = - {\csc}^{2} x - \cos x + {\sec}^{2} x$

Our possible critical points occur where $f \left(x\right)$ is defined, but $f ' \left(x\right)$ either equals 0 or is undefined. First we check for undefined points: in this situation these will occur whenever we must divide by 0. Since $\csc x = \frac{1}{\sin} x$ and $\sec x = \frac{1}{\cos} x$, these occur when either $\sin \left(x\right)$ or $\cos \left(x\right)$ is equal to 0. These occur at the same points we discovered above, the iterations of $n \frac{\pi}{2}$. We know these cannot be critical points, so there are no critical points occurring where the derivative is undefined.

Next we will check for where $f ' \left(x\right) = 0$. At this point it may be beneficial to utilize a graphing calculator; none of the calculations I have run without a calculator have helped to find the roots. Provided below is the appropriate graph of the derivative; the zeroes of this function between $\left[0 , 2 \pi\right]$ are the desired critical points.

graph{-(csc(x))^2 - cos(x) + (sec(x))^2 [-0.1, 6.3, -10, 10]}