91Ó°ÊÓ

When working with functions, it's crucial to identify the domain, which is the set of all possible input values (x-values) for the function. Imagine a function being like a machine. You need to know what you can input into the machine without breaking it. For rational functions like \( f(x) = \frac{ax^2}{x-b} \), you need to ensure the denominator isn't zero, as division by zero is undefined.
In this problem, the expression \( x - b \) in the denominator cannot be zero. Thus, \( x eq b \).
The domain of \( f \) is all real numbers except for \( x = b \). To make this clear, we often write this domain in interval notation as \((-\infty, b) \cup (b, \infty)\). This tells us that the function works for every x-value except for the danger zone, \( x = b \).
Understanding domains is not only essential for finding valid solutions but also for anticipating any potential issues when analyzing function behaviors.

Finding a derivative of a quotient might seem complex, but the quotient rule makes it manageable. It's like having a formula ready to tackle functions divided by each other (hence the name, quotient). The quotient rule states: if you have a function \( h(x) = \frac{u(x)}{v(x)} \), then the derivative \( h'(x) \) is given by:
\[ h'(x) = \frac{u'v - uv'}{v^2} \]
In the context of our problem, where \( f(x) = \frac{ax^2}{x-b} \), we identify \( u(x) = ax^2 \) and \( v(x) = x-b \).

• Compute \( u' = 2ax \)
• Compute \( v' = 1 \)

Substitute these into the quotient rule formula to find the derivative \( f'(x) \). The result is:
\[ f'(x) = \frac{(2ax)(x-b) - (ax^2)(1)}{(x-b)^2} \]
By simplifying, we arrive at:
\[ f'(x) = \frac{ax(x-2b)}{(x-b)^2} \].
The quotient rule makes finding derivatives of fractions straightforward, allowing us to handle more intricate calculus problems with ease.

Derivatives are powerful tools in calculus, telling us how a function changes as its input changes. They are the key to understanding various properties of functions, including identifying critical points. Critical points occur where the derivative is zero or undefined, which often leads to insights about the graph's behavior, such as maximums, minimums, or points of inflection.
To find where \( f(x) = \frac{ax^2}{x-b} \) has critical points, we first found its derivative using the quotient rule: \( f'(x) = \frac{ax(x-2b)}{(x-b)^2} \). To locate critical points, set the derivative \( f'(x) \) equal to zero and solve:

• Focus on the numerator, as zeroing any fraction equates to the numerator being zero.

Hence, solve \( ax(x - 2b) = 0 \), which simplifies to \( x = 0 \) or \( x = 2b \). The function \( f'(x) \) becomes undefined at \( x = b \) because the denominator \((x-b)^2\) would be zero, but \( x = b \) isn't in the function's domain anyway.
By discovering where the derivative is zero, we identified the critical points at \( x = 0 \) and \( x = 2b \). This indicates positions where the function takes a halt in rising or falling, potentially corresponding to local maximum or minimum points.