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The quotient rule is a crucial tool in calculus, especially when finding the derivative of a function that is the ratio of two differentiable functions. When you have a function expressed as a quotient like \(f(x) = \frac{u(x)}{v(x)}\), the quotient rule helps you compute its derivative. The formula for the quotient rule is:

• \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\)

Here, \(u\) and \(v\) are differentiable functions of \(x\). In our example exercise, \(u = 5x^2 - 18x + 45\) and \(v = x^2 - 9\). To apply the quotient rule correctly, find the derivatives \(u'\) and \(v'\), and then insert these into the formula. This will help you obtain \(f'(x)\), the derivative of the original function. It's essential to remember that the quotient rule requires both the numerator and the denominator to be differentiable.

Derivative computation is about finding how a function changes at any point and is key to identifying critical points. The derivative, represented as \(f'(x)\), provides a rate of change that can be zero at specific points called critical points. In the original exercise, once you use the quotient rule to compute the derivative of \(f(x)\), you will aim to solve for \(x\) when \(f'(x) = 0\).

• This means setting the numerator of the obtained derivative formula to zero because the fraction is zero only when its numerator is zero.

Performing the needed algebra will lead you to the potential critical points. These are the \(x\) values where \(f'(x)\) changes sign, indicating maximum, minimum, or inflection points in the graph of \(f(x)\). Don't forget to consider scenarios where \(f'(x)\) is undefined, as these also contribute to your set of critical points.

Understanding the domain of a function is fundamental when looking for critical points because the domain determines where the function is valid and defined. For rational functions like \(f(x) = \frac{5x^2 - 18x + 45}{x^2 - 9}\), the domain is all real numbers except where the denominator equals zero. Solving \(x^2 - 9 = 0\) gives us the values \(x = 3\) and \(x = -3\). Hence, the domain of \(f(x)\) is all real numbers except \(-3\) and \(3\).

• Critical points derived from your derivative computations must lie within this function domain, or else they are not valid for your original function.

Always cross-check that each critical point satisfies the domain of the function. This is crucial when you determine if a point is indeed a critical point by the definition given in the context of the exercise.

In calculus, the derivative is "undefined" at points where it does not exist, often due to division by zero or other constraints. In the evaluated function \(f(x)\), \(f'(x)\) is undefined wherever the original denominator \(x^2 - 9\) is zero. This happens at \(x = 3\) and \(x = -3\).

• These points often correspond to vertical asymptotes in graphical analysis and are important when analyzing function behavior.

Identifying undefined derivatives helps in understanding the full behavior of the function along with its valid range. While such points make \(f'(x)\) undefined, they aren't considered critical points since they are not in the domain of the function. This is a critical distinction because valid critical points must lie both where the derivative is zero or undefined and within the domain.