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Derivative Calculus involves computing the derivative of a function, which gives the rate at which the function's value is changing at any given point. To find the derivative of a function like \(g(x) = x \sqrt{9-x}\), we use rules such as the product rule and the chain rule.
These rules help us differentiate complex expressions systematically.The **product rule** states that if you have a product of two functions, say \(u\) and \(v\), then the derivative \((uv)'\) is \(u'v + uv'\). In our example, \(u = x\) and \(v = \sqrt{9-x}\). The **chain rule** is used when dealing with composite functions, like \(\sqrt{9-x}\), where the outer function is the square root, and the inner function is \(9-x\).
By applying these rules, we calculate the derivative \(g'(x) = \frac{18 - 3x}{2 \sqrt{9 - x}}\). This derivative tells us how the function \(g(x)\) changes with \(x\).

Finding critical points is about identifying where a function reaches its relative extremes—maximum or minimum values.
Critical points occur where the derivative of a function is zero or undefined.To find the critical points of \(g(x) = x \sqrt{9-x}\), we set its derivative \(g'(x)\) to zero. The equation \(\frac{18 - 3x}{2 \sqrt{9-x}} = 0\) simplifies to \(18 - 3x = 0\), giving us \(x = 6\).
It's crucial to check whether this value falls within the domain of the function, which is \(0 \leq x \leq 9\). Since it does, \(x=6\) is a valid critical point.

The First Derivative Test is used to determine whether a critical point is a relative maximum, a relative minimum, or neither. This test examines the sign of the derivative before and after the critical point.For the function \(g(x) = x \sqrt{9-x}\), we saw that \(x=6\) is a critical point. Before \(x=6\), the derivative \(g'(x)\) is positive, indicating the function is increasing. After \(x=6\), the derivative is negative, showing the function is decreasing.
Therefore, according to the First Derivative Test, \(x=6\) is a point of relative maximum. This test provides a straightforward way to determine the behavior of the function around critical points.

The Second Derivative Test helps to determine concavity and identify points of inflection in a function. It provides more information about the curvature of the graph.To perform this test, we need the second derivative of the function. For \(g(x) = x \sqrt{9-x}\), we calculate \(g''(x) = \frac{27(4x-18)}{4(9-x)^{3/2}}\).
Setting \(g''(x) = 0\), we find \(x=4.5\). This means the concavity changes at this point, indicating \(x=4.5\) is a point of inflection. Around this point, checking the changes in the sign of \(g''(x)\) confirms a shift in concavity, affirming \(x=4.5\) as where the graph switches from being concave up to concave down or vice versa.