Comnpletely (Algebraically) analyze

explanation

\(\blacktriangleright \) Category

\(R(y) = -3 \sqrt {5-2y} + 4\) matches our official template, \(A \sqrt [n]{B \, x + C} + D \), which makes it a (even) radical function.

\(\blacktriangleright \) Domain

\(R(y)\) is an even root, which means its natural domain is all real numbers that make the inside nonegative.

The inside is \(5-2y\), which is a linear function with a negative leading coefficient. That tells us that it is nonnegative for numbers less than its zero. Its zero is \(\frac {5}{2}\).

The domain of \(R\) is \(\left ( -\infty , \frac {5}{2} \right ]\)

\(\blacktriangleright \) Zeros

\begin{align*} -3 \sqrt {5-2y} + 4 &= 0 \\ \sqrt {5-2y} &= \frac {4}{3} \\ 5 - 2y &= \frac {16}{9} \\ 5 - \frac {16}{9} &= 2y \\ \frac {1}{2} \left ( 5 - \frac {16}{9} \right ) &= y \end{align*}

\(\frac {1}{2} \left ( 5 - \frac {16}{9} \right ) \approx 1.6111\), which agrees with the graph.

\(\blacktriangleright \) Continuity

\(R\) is continuous, since it is a root function and all root funcitons are continuous.

\(\blacktriangleright \) End-Behavior

As an even root, \(R\) only has end-behavior in one direction. It is unbounded as \(y\) approaches \(-\infty \).

Since the leading coefficient is \(-3 < 0\), R is unbounded negatively.

\(\blacktriangleright \) Behavior

A root or radical function is always increasing or always decreasing.

The Chain Rule will tell us which.

We are viewing \(R\) as a composition.

• Let \(L_{out}(v) = -3 \, v + 4\), a decreasing linear function
• Let \(SQ(t) = \sqrt {t}\), an increasing core square root function
• Let \(L_{in}(u) = 5 - 2 \, u\), a decreasing linear function

The Chain Rule tells us that \(R\) is an increasing function.

\(\blacktriangleright \) Global Maximum and Minimum

The end-behavior tells us that there is no global minimum. And, since \(R\) is always increasing, its global maximum value must occur at \(\frac {5}{2}\)

\(\blacktriangleright \) Local Maximum and Minimum

As a radical function, \(R\) has a single local maximum or minimum, which is the same as the global maximum or minimum.

Therefore, the only local maximum is \(4\), which occurs at \(\frac {5}{2}\)

\(\blacktriangleright \) Range

• \(R\) is continuous
• \(\lim \limits _{y \to -\infty } R(y) = -\infty \)
• The global maximum is \(4\)

The range of \(R\) is \((-\infty , 4]\)

\(\blacktriangleright \) Graph

The graph of \(z = R(y) = -3 \sqrt {5-2y} + 4\)