Completely analyze \(K(w) = w \sqrt {9-w}\) with \(K'(w) = \sqrt {9 - w} + w \frac {1}{2} (9 - w)^{-\tfrac {1}{2}} (-1)\)

\(K\) is not an elementary function. It is the product of a linear function and a root function.

Domain

The domian of \(K\) consists of the real numbers that are included in both domains of both factor functions.

\(w\) is a linear function, so its domain is \((-\infty , \infty )\).

\(\sqrt {9-w}\) is a root function. The inside must be nonnegative. The decide this, we first find the zero of \(9 - w\), which is \(9\).

Since \(9 - w\) is a linear function with a negative leading coefficient, \(9 - w > 0\) on \((-\infty , 9)\).

Therefore, ther domain of \(K\) is \((-\infty , 9]\).

Zeros

\(K\) is already written in factored form, so we will use the zero product property.

Either \(w = 0\) or \(\sqrt {9-w} = 0\).

Either \(w = 0\) or \(w = 9\).

\(K\) has two zeros, \(0\) and \(9\).

Continuity

LInear and root functions are both continuous, so is there product.

End-Behavior

We only have one direction for end-behavior.

Since \(w\) is a linear function with a positive leading coefficient, we know that \(\lim \limits _{w \to -\infty } w = -\infty \). It is unbounded negatively.

Since \(\sqrt {9-w}\) is a root funciton with a positive leading coefficient, we know that \(\lim \limits _{w \to -\infty } \sqrt {9-w} = \infty \). It is unbounded negatively.

The product of two functions which are both unbounded is unbounded. Since one takes on negative values and the ohter postive values, their product is negative.

Behavior (Increasing and Decreasing)

We’ll use the derivative.

\(6\) is a domain number and a zero of the derivative. That makes \(6\) a critical number.

\(9\) is a domain number where the derivative does not exist. That makes \(9\) a critical number.

\(K'(w)\) is a product. The root factor is always positive, so the sign of \(K'(w)\) is the same as the sign of \(6 - w\).

\(6 - w\) is a linear function with a negative leading coeffcient. So, \(6 - w > 0\) on \((-\infty , 6)\) and \(6 - w < 0\) on \((6, 9)\).

\(\blacktriangleright \) \(K\) is increasing on \((-\infty , 6)\)

\(\blacktriangleright \) \(K\) is decreasing on \((6, 9)\)

Global Maximum and Minimum

Since \(\lim \limits _{w \to -\infty } K(w) = -\infty \), there is no global minimum.

Since \(K\) is increasing on \((-\infty , 6)\) and \(K\) is decreasing on \((6, 9)\), \(K\) has a global maximum at \(6\).

The global maximum value is \(K(6) = 6 \sqrt {9 - 6} = 6 \sqrt {3}\).

Local Maximum and Minimum

Since global maximums are automatically local maximums, \(6 \sqrt {3}\) is a local maximum.

We also have the critical number \(9\) with \(K(9) = 0\). This is a local minimum. We can see this because \(K\) is contiuous and decreasing on the interval \((6, 9]\).

Range

• \(K\) is continuous
• \(6 \sqrt {3}\) is the global maximum
• \(\lim \limits _{w \to -\infty } K(w) = -\infty \)

That gives a range of \((-\infty , 6 \sqrt {3}]\).

This all agrees with the graph.

\(\blacktriangleright \) desmos graph