Before we start, it behooves us to analyze and see if sin xsinx takes on a value of 00 at any point on the interval. sin xsinx is zero for all x such that x = npix=nπ. pi/2π2 and 3pi/43π4 are both less than piπ and greater than 0pi = 00π=0; thus, sin xsinx 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)
