91Ó°ÊÓ

In calculus, the derivative of a function measures how the function value changes as its input changes. For the provided function, \(f(x) = \frac{\sin x}{3 - \sin x}\), finding the derivative is essential to locate the critical points where the function could reach its absolute maximum or minimum values.
To find the derivative of this function, we use the quotient rule, which is a method for differentiating functions that are ratios of two differentiable functions.
The quotient rule states: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
Given our function: \(u = \sin x\) and \(v = 3 - \sin x\), with the derivatives \(u' = \cos x\) and \(v' = -\cos x\), we get: \[ f'(x)=\frac{\cos x (3 - \sin x) - \sin x (-\cos x)}{(3 - \sin x)^2} = \frac{3 \cos x}{(3 - \sin x)^2} \].
This expression will be used to find the critical points, which are essential for determining the absolute maxima and minima.

Critical points are where the derivative of a function is either zero or undefined. These points are potential candidates for local and global maxima and minima.
To find the critical points for our function, we need to solve for when the derivative equals zero:

• Start with the derivative: \(f'(x) = \frac{3 \cos x}{(3 - \sin x)^2}\)
• Set the derivative equal to zero: \(0 = \frac{3 \cos x}{(3 - \sin x)^2}\)
• The fraction's numerator must be zero, so solve: \(3 \cos x = 0 \implies \cos x = 0\)

On the interval \([0, 2\pi]\), the cosine function is zero at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\). These are our critical points, where the function may reach maximum or minimum values.

The quotient rule is specifically used for finding the derivative of the division of two functions. Remember the rule: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \].
For our function \(f(x) = \frac{\sin x}{3 - \sin x}\), here's a step-by-step application:

• Identify \(u\) and \(v\): \(u = \sin x\) and \(v = 3 - \sin x\)
• Find their derivatives: \(u' = \cos x\) and \(v' = -\cos x\)
• Apply the quotient rule: \[ f'(x)=\frac{\cos x (3 - \sin x) - \sin x (-\cos x)}{(3 - \sin x)^2} = \frac{3 \cos x}{(3 - \sin x)^2}\]

The quotient rule is very handy for functions presented as the ratio of two expressions, easing the process of differentiation.

The evaluation of a function at specific points helps in determining the actual values of the function, especially at critical points and endpoints to find the maxima and minima.
For \(f(x) = \frac{\sin x}{3 - \sin x}\), evaluate the function at critical points and the interval boundaries:

• At \(x=0\): \ \[f(0) = \frac{\sin 0}{3 - \sin 0} = \frac{0}{3 - 0} = 0\]
• At \(x = \frac{\pi}{2}\): \ \[f\left(\frac{\pi}{2}\right) = \frac{\sin\left(\frac{\pi}{2}\right)}{3 - \sin\left(\frac{\pi}{2}\right)} = \frac{1}{3 - 1} = \frac{1}{2}\]
• At \(x = \frac{3\pi}{2}\): \ \[f\left(\frac{3\pi}{2}\right) = \frac{\sin\left(\frac{3\pi}{2}\right)}{3 - \sin\left(\frac{3\pi}{2}\right)} = \frac{-1}{3 - (-1)} = -\frac{1}{4}\]
• At \(x = 2\pi\): \ \[f(2\pi) = \frac{\sin(2\pi)}{3 - \sin(2\pi)} = \frac{0}{3 - 0} = 0\]

These evaluations demonstrate that the function attains its absolute maximum of \(\frac{1}{2}\) at \(x = \frac{\pi}{2}\) and its absolute minimum of \(-\frac{1}{4}\) at \(x = \frac{3\pi}{2}\).