What are the extrema of f(x)=3x-1/sinx  on [pi/2,(3pi)/4]?

Nov 13, 2015

The absolute minimum on the domain occurs at approx. $\left(\frac{\pi}{2} , 3.7124\right)$, and the absolute max on the domain occurs at approx. $\left(3 \frac{\pi}{4} , 5.6544\right)$. There are no local extrema.

Explanation:

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 $x = n \pi$. $\frac{\pi}{2}$ and $3 \frac{\pi}{4}$ are both less than $\pi$ and greater than $0 \pi = 0$; thus, $\sin x$ does not take on a value of zero here.

In order to determine this, recall that an extreme occurs either where $f ' \left(x\right) = 0$ (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

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(3 x\right) - \frac{d}{\mathrm{dx}} \left(\frac{1}{\sin} x\right) = 3 - \frac{d}{\mathrm{dx}} \left(\frac{1}{\sin} x\right)$

How should we solve this last term?

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

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

when $g \left(x\right) \ne 0$

Returning to our main equation, we left off with;

$3 - \frac{d}{\mathrm{dx}} \left(\frac{1}{\sin} x\right)$.

Since $\sin \left(x\right)$ is differentiable, we can apply the reciprocal rule here:

$3 - \frac{d}{\mathrm{dx}} \left(\frac{1}{\sin} x\right) = 3 - \frac{- \cos x}{\sin} ^ 2 x$

Setting this equal to 0, we arrive at:

$3 + \cos \frac{x}{\sin} ^ 2 x = 0.$

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

$\cos \frac{x}{\sin} ^ 2 x = - 3 \implies \cos \frac{x}{1 - {\cos}^{2} x} = - 3 \implies \cos x = - 3 + 3 {\cos}^{2} x \implies 3 {\cos}^{2} x - \cos x - 3 = 0$

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

$3 {u}^{2} - u - 3 = 0 = a {u}^{2} + b u + c$. Using the quadratic formula here...
$\frac{1 \pm \sqrt{1 - 4 \left(- 9\right)}}{6} = \frac{1 \pm \sqrt{37}}{6}$

Our roots occur at $u = \frac{1 \pm \sqrt{37}}{6}$ according to this. However, one of these roots ($\frac{1 + \sqrt{37}}{6}$) cannot be a root for $\cos x$ because the root is greater than 1, and $- 1 \le \cos x \le 1$ 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 $f \left(\frac{\pi}{2}\right) \approx 3.7124 , f \left(3 \frac{\pi}{4}\right) \approx 5.6544$

Thus, our absolute minimum on the domain is approximately $\left(\frac{\pi}{2} , 3.7124\right) ,$ and our maximum is approximately $\left(3 \frac{\pi}{4} , 5.6544\right)$