Describe Everything

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Complete Analysis

Completely analyze

\[ B(f) = \frac {(f+3)(2f-1)}{(f+5)(f-2)} \]

with

\[ B'(f) = \frac {f^2 - 34f - 41}{(f+5)^2(f-2)^2} \]

Categorize: $B(f)$ is a rational function, because it is of the form $\frac {polynomial}{polynomial}$

Domain

The natural domain of a rational function is all real numbers except the zeros of the denominator, which are $-5$ and $2$.

The domain is $(-\infty , -5) \cup (-5, 2) \cup (2, \infty )$.

Zeros

The zeros of a rational function are the zeros of the numerator, which are not also zeros of the denominator, i.e. they are in the domain.

$-3$ and $\answer {\frac {1}{2}}$ are the zeros of $B$.

Continuity

Rational functions are continuous.

Rational functions can have singularities. Here the zeros of the denominator are $-5$ and $2$. These are not zeros of the numerator. That makes them asymptotic singularities. So, we know $B$ is unbounded near them. We just need to figure out the sign.

Let’s map out the signs of $B$.

$B$ has four zeros and/or singularities. In order, they are $-5$, $-3$, $\frac {1}{2}$, and $2$.

They all have multiplicity $1$, which is odd. That means $B$ switches signs across them.

The end-behavior will get us started on the signs.

End-Behavior

$B$ is a rational function.

The degree of the numerator is $\answer {2}$.

The degree of the denominator is $\answer {2}$.

$B$ is a rational function and the degree of the denominator is greater than less than equal to the degree of the numerator.

Since the degrees of the numerator and denominator are equal, the end-behvior is the quotient of the leading coefficients.

\[ \lim _{f \to -\infty } B(f) = \frac {2}{1} = 2 \, \text { and } \, \lim _{f \to \infty } B(f) = \frac {2}{1} = 2 \]

Continuing with singularity behavior....

We now know that $B$ is positive for large negative domain numbers. This switches at $-5$, which is the first singularity.

$B$ is positive on $(-\infty , -5)$

That tells us that

\[ \lim \limits _{f \to -5^-} B(f) = \infty \]

The sign of $B$ switches at $-5$, which gives us

\[ \lim \limits _{f \to -5^+} B(f) = -\infty \]

$B$ switches signs at $-3$, which is the first zero.

Therefore, $B$ is negative on $(-5, -3)$. Then, $B$ switches to positive on $\left ( -3, \frac {1}{2} \right )$ Then, $B$ switches to negative on $\left ( \frac {1}{2}, 2 \right )$.

That tells us that

\[ \lim \limits _{f \to 2^-} B(f) = -\infty \]

Then, $B$ switches to positive on $(2, \infty )$.

\[ \lim \limits _{f \to 2^+} B(f) = \infty \]

Behavior (Increasing and Decreasing)

We’ll use the derivative to help us with behavior.

\[ B'(f) = \frac {f^2 - 34f - 41}{(f+5)^2(f-2)^2} \]

$B'(f)$ is also a rational function. So, it is continous. It has the same singularities as $B$. But it has different zeros.

To get the critical numbers, we need to factor the numerator.

\[ f = \frac {34 \pm \sqrt {(-34)^2 - 4 (1)(-41)}}{2(1)} = \frac {34 \pm \sqrt {1320}}{2} = \frac {34 \pm 2\sqrt {330}}{2} = 17 \pm \sqrt {330} \]

Let’s check with a graph.

$17 - \sqrt {330} \approx -1.165902125$

$17 + \sqrt {330} \approx 35.16590212$

$\blacktriangleright $ desmos graph

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$\blacktriangleright $ desmos graph

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It seems our algebra is good.

\[ B'(f) = \frac {(f - (17 - \sqrt {330}))(f - (17 + \sqrt {330}))}{(f+5)^2(f-2)^2} \]

$B'$ has two zeros and two singularities.

\[ 17 - \sqrt {330}, 17 + \sqrt {330}, -5, 2 \]

We need their order.

The graph suggests that $-5 < 17 - \sqrt {330}$. Let’s see if we can show that is true.

\[ -5 < -1 = 17 - 18 = 17 -\sqrt {324} < 17 - \sqrt {330} \]

Since $ \sqrt {330} > \sqrt {289} = 17$, we know that $17 - \sqrt {330} < 0$.

We also know that $2 < 17 + \sqrt {330} $.

That gives us

\[ -5 < 17 - \sqrt {330} < 2 < 17 + \sqrt {330} \]

$17 - \sqrt {330}$ and $17 + \sqrt {330}$ are zeros of $B'$ with multiplicity $1$, which is odd. $B'$ will change signs across them.

$-5$ and $2$ are singularities of $B'$ with multiplicity $2$, which is even. $B'$ will not change signs across them.

We just need a starting sign.

For very large negative numbers (anything less than $17 - \sqrt {330}$),

\[ B'(f) = \frac {negative \cdot negative}{positive \cdot positive} = positive \]

Therefore, the signs of $B'$ are

$B'$ is positive on $(-\infty , -5)$
$B'$ is positive on $(-5, 17 - \sqrt {330})$
$B'$ is negative on $(17 - \sqrt {330}, 2)$
$B'$ is negative on $(2, 17 + \sqrt {330})$
$B'$ is positive on $(17 + \sqrt {330}, \infty )$

This give us the behavior of $B$.

$B$ is increasing on $(-\infty , -5)$
$B$ is increasing on $(-5, 17 - \sqrt {330})$
$B$ is decreasing on $(17 - \sqrt {330}, 2)$
$B$ is decreasing on $(2, 17 + \sqrt {330})$
$B$ is increasing on $(17 + \sqrt {330}, \infty )$

Global Maximums and Minimums

$B$ has no global maximum or minimum, because

\[ \lim \limits _{f \to 2^-} B(f) = -\infty \]

\[ \lim \limits _{f \to 2^+} B(f) = \infty \]

Local Maximums and Minimums

$B$ has two critical numbers. They are locations for local maximums or minimums.

$B$ is increasing on $(-5, 17 - \sqrt {330})$
$B$ is decreasing on $(17 - \sqrt {330}, 2)$

Therefore, $B(17 - \sqrt {330})$ is a local maximum occurring at $17 - \sqrt {330}$.

$B$ is decreasing on $(2, 17 + \sqrt {330})$
$B$ is increasing on $(17 + \sqrt {330}, \infty )$

Therefore, $B(17 + \sqrt {330})$ is a local minimum occurring at $17 + \sqrt {330}$.

Range

To figure out the range, we need to know if there is a gap. We need to know if $B(17 + \sqrt {330})$ or $B(17 - \sqrt {330})$ is greater.

The graph suggests that

\[ B(17 + \sqrt {330}) > B(17 - \sqrt {330}) \]

That is going to be a bit of algebra. Let’s ask Wolfram Alpha to simply each of those.

\[ B(17 - \sqrt {330}) = \frac {1}{49} (61 - 2 \sqrt {330}) \]

\[ B(17 + \sqrt {330}) = \frac {1}{49} (61 + 2 \sqrt {330}) \]

Confirmed.

\[ B(17 + \sqrt {330}) > B(17 - \sqrt {330}) \]

There is a gap in the range.

The range of $B$ is

\[ \left ( -\infty , \frac {1}{49} (61 - 2 \sqrt {330}) \right ] \cup \left [ \frac {1}{49} (61 + 2 \sqrt {330}), \infty \right ) \]

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2025-08-25 18:10:23