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

1 Answer
Dec 27, 2015

Answer:

There are no critical points at any of the #npi/2# values; #f(x)# 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(x)# occurs where the derivative of #f(x)#, denoted as either #f'(x)# or #(df)/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 = (cos x)/sin x# and #tan x = (sin x)/cos x#.

Since one cannot divide by zero, the original function is discontinuous in the neighborhood of those values for which #sin (x)# or #cos (x) = 0#. This consists of any value which can be defined as #npi/2#, where #n# is an integer. Within our domain we have five such values: #0, pi/2, 2pi/2 = pi, 3pi/2, 4pi/2 = 2pi#. 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.

#d/dx tan(x) = d/dx (sin x/cosx) = (cos^2(x)+sin^2(x))/cos^2x = 1/cos^2x = sec^2x#
#d/dx cot (x) = d/dx (cos x/sin x) = (-sin^2x - cos^2x)/sin^2x = -1(sin^2x + cos^2x)/sin^2x = -1/sin^2x = -csc^2x#

Thus, differentiating our original #f(x)# yields #d/dx (cot x - sin x + tan x) = -csc^2x - cos x + sec ^2x#

Our possible critical points occur where #f(x)# is defined, but #f'(x)# 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 = 1/sinx# and #sec x = 1/cos x#, these occur when either #sin (x)# or #cos (x)# is equal to 0. These occur at the same points we discovered above, the iterations of #npi/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'(x) = 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 #[0,2pi]# are the desired critical points.

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