Before we start, it behooves us to analyze and see if #sin x# takes on a value of #0# at any point on the interval. #sin x# is zero for all x such that #x = npi#. #pi/2# and #3pi/4# are both less than #pi# and greater than #0pi = 0#; thus, #sin x# does not take on a value of zero here.

In order to determine this, recall that an extreme occurs either where #f'(x) = 0# **(critical points)** or at one of the endpoints. This in mind, we take the derivative of the above f(x), and find points where this derivative equals 0

#(df)/dx = d/dx (3x) - d/dx (1/sin x) = 3 - d/dx (1/sinx)#

How should we solve this last term?

Consider briefly the **reciprocal rule**, which was developed to handle situations such as our last term here, #d/(dx) (1/sin x)#. The reciprocal rule allows us to bypass directly using the chain or quotient rule by stating that given a differentiable function #g(x)#:

#d/dx 1/g(x) = (-g'(x))/((g(x))^2#

when #g(x)!= 0 #

Returning to our main equation, we left off with;

#3 - d/dx (1/sin x)#.

Since #sin(x)# is differentiable, we can apply the reciprocal rule here:

#3 - d/dx (1/sin x) = 3 - (-cos x)/sin^2x#

Setting this equal to 0, we arrive at:

#3 + cos x/sin^2x = 0. #

This can only occur when #cos x / sin^2 x = -3.#. From here it may behoove us to use one of the trigonometric definitions, specifically #sin^2x = 1 - cos^2 x#

#cosx/sin^2x = -3 => cosx/(1-cos^2x) = -3 =>cos x = -3 + 3cos^2x => 3cos^2x - cos x - 3 = 0#

This resembles a polynomial, with #cos x# replacing our traditional x. Thus, we declare #cos x = u# and...

#3u^2 - u - 3 = 0 = au^2 + bu +c#. Using the quadratic formula here...

#(1 +- sqrt(1 - 4(-9)))/6 = (1+- sqrt 37)/6#

Our roots occur at #u = (1+-sqrt37)/6# according to this. However, one of these roots (#(1 + sqrt37)/6#) cannot be a root for #cos x# because the root is greater than 1, and #-1<=cosx<=1# for all x. Our second root, on the other hand, calculates as approximately #-.847127#. However, this is less than the minimum value the #cos x# function can on the interval (since #cos (3pi/4) = -1/sqrt 2) = -.707 < -.847127#. Thus, *there is no critical point in the domain*.

This in mind, we must return to our endpoints and put them into the original function. Doing so, we obtain #f(pi/2) approx 3.7124, f(3pi/4) approx 5.6544#

Thus, our absolute minimum on the domain is approximately #(pi/2, 3.7124),# and our maximum is approximately # (3pi/4, 5.6544)#