Analyzing

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describe everything

$\blacktriangleright $ Reasoning: Reasoning is a logical explanation that describes our conclusions, how we arrived at those conclusions, and why we think those conclusions are correct.

Analysis is not a list of conclusions. We are not looking for such a list.

We are looking for the thought process that arrived at the list of conclusions.

Completely analyze

\[ R(t) = \frac {(t+3)(t+1)}{(t-2)(t-3)} \]

with

\[ R'(t) = -\frac {9 t^2 -6 t -39}{(t-2)^2 (t-3)^2} \]

$R(t)$ is not a composition. It is a rational function.

Here is a quick DESMOS graph.

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There is something wrong with this graph. $2$ and $3$ make the denominator $0$, but the numbers in the interval $(2, 3)$ are in the domain. There should be a piece of the graph there that is not showing up.

Let’s evalaute at $2.5$ and see what type of values we are getting.

$R(2.5) = -77$

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There is the other part of the graph.

From the graph, we are anticipating two singularities, two critical numbers, switching between increasing and decreasing, and end-behavior which is not unbounded.

$\blacktriangleright $ Domain

$R(t)$ is a rational function. Its domain is all real numbers except the zeros of the denominator.

The domain is $(-\infty , 2) \cup (2,3) \cup (3, \infty )$

$\blacktriangleright $ Zeros

$R(t)$ is a rational function. Its zeros are the zeros of the numerator except those that are also zeros of the denominator.

The zeros of $R$ are $-3$ and $-1$.

$\blacktriangleright $ Continuity

$R(t)$ is a rational function. It is continuous.

$R$ does have singularities. We need the behavior around them. Both $2$ and $3$ are zeros of the denominator and not zeros of the numerator. This makes them asymptotic singularities.

We knw know that $R$ is unbounded as $t$ approaches $2$ and $3$. We need to figure out the sign.

For $\lim \limits _{t \to 2^-}R(t)$, $t < 2$, which gives

$t + 3 > 0$
$t + 1 > 0$
$t - 2 < 0$
$t - 3 < 0$

The sign of $R$ is $\frac {pos \cdot pos}{neg \cdot neg} = positive$

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

For $\lim \limits _{t \to 2^+}R(t)$, $t > 2$, which gives

$t + 3 > 0$
$t + 1 > 0$
$t - 2 > 0$
$t - 3 < 0$

The sign of $R$ is $\frac {pos \cdot pos}{pos \cdot neg} = negative$

\[ \lim \limits _{t \to 2^+}R(t) = -\infty \]

For $\lim \limits _{t \to 3^-}R(t)$, $t < 3$, which gives

$t + 3 > 0$
$t + 1 > 0$
$t - 2 > 0$
$t - 3 < 0$

The sign of $R$ is $\frac {pos \cdot pos}{pos \cdot neg} = negative$

\[ \lim \limits _{t \to 3^-}R(t) = -\infty \]

For $\lim \limits _{t \to 3^+}R(t)$, $t > 3$, which gives

$t + 3 > 0$
$t + 1 > 0$
$t - 2 > 0$
$t - 3 > 0$

The sign of $R$ is $\frac {pos \cdot pos}{pos \cdot pos} = positive$

\[ \lim \limits _{t \to 3^+}R(t) = \infty \]

This agrees with the graph.

$\blacktriangleright $ End-Behavior

$R(t)$ is a rational function. The degree of the numerator and the degree of the denomiator are equal. The end-behavior is the quotient is the leading coefficients.

\[ \lim \limits _{t \to -\infty }R(t) = \frac {1}{1} = 1 \]

\[ \lim \limits _{t \to \infty }R(t) = \frac {1}{1} = 1 \]

$\blacktriangleright $ Behavior Rate-of-Change Increasing and Decreasing

The sign of $R'(t)$ will give us the behavior of $R(t)$.

To get the sign of $R'(t)$, we will factor it.

The denominator is factored. The numerator is a quadratic function. To factor it, we need its zeros, which we can get from the quadratic formula.

But, first we might as well factor out $3$ in the numerator.

\[ R'(t) = -\frac {9 t^2 -6 t -39}{(t-2)^2 (t-3)^2} = -\frac {3 (3 t^2 - 2 t - 13)}{(t-2)^2 (t-3)^2} \]

$3 (3 t^2 - 2 t - 13) = 0$

$3 t^2 - 2 t - 13 = 0$

\[ t = \frac {2 \pm \sqrt {(-2)^2 - 4(3)(-13)}}{2(3)} = \frac {2 \pm \sqrt {160}}{6} = \frac {2 \pm 4 \sqrt {10}}{6} = \frac {1 \pm 2 \sqrt {10}}{3} \]

Two critical numbers: $\frac {1 - 2 \sqrt {10}}{3}$ and $\frac {1 + 2 \sqrt {10}}{3}$

Quick Check:

\[ \frac {1 - 2 \sqrt {10}}{3} \approx -1.774851773 \]

\[ \frac {1 + 2 \sqrt {10}}{3} \approx 2.44151844 \]

These agree with the graph, so we feel like our algebra is correct.

We can now factor $R'(t)$

\[ R'(t) = -\frac {3 \left (t - \left (\frac {1 - 2 \sqrt {10}}{3} \right ) \right ) \left (t - \left (\frac {1 + 2 \sqrt {10}}{3} \right ) \right ) }{ (t-2)^2 (t-3)^2 } \]

Both $\frac {1 - 2 \sqrt {10}}{3}$ and $\frac {1 - 2 \sqrt {10}}{3}$ have odd multiplicities, so $R'$ will change signs across them.

Both $2$ and $3$ have even multiplicities, so $R'$ will not change signs across them.

We need the order of $\frac {1 - 2 \sqrt {10}}{3}$, $\frac {1 - 2 \sqrt {10}}{3}$, $2$, and $3$.

That graph makes us think that

\[ \frac {1 - 2 \sqrt {10}}{3} < 2 < \frac {1 + 2 \sqrt {10}}{3} < 3 \]

Let’s show that $\frac {1 - 2 \sqrt {10}}{3} < 2$.

This true because $\frac {1 - 2 \sqrt {10}}{3} < 0$, since $1 < 2 \sqrt {10}$.

Next, we’ll show that $2 < \frac {1 + 2 \sqrt {10}}{3}$.

This is true if $6 < 1 + 2 \sqrt {10}$, which is true since $3 < \sqrt {10}$.

Finally, we need to show that $\frac {1 + 2 \sqrt {10}}{3} < 3$

This is true if $1 + 2 \sqrt {10} < 9$, which is true since $\sqrt {10} < 4$.

We now know the ordering of the critical numbers and singularities.

\[ \frac {1 - 2 \sqrt {10}}{3} < 2 < \frac {1 + 2 \sqrt {10}}{3} < 3 \]

On $\left ( -\infty , \frac {1 - 2 \sqrt {10}}{3} \right )$,

$t < \frac {1 - 2 \sqrt {10}}{3}$ and $\left ( t - \frac {1 - 2 \sqrt {10}}{3} \right ) < 0$
$t < 2$ and $(t - 2) < 0$
$t < \frac {1 + 2 \sqrt {10}}{3}$ and $\left ( t - \frac {1 + 2 \sqrt {10}}{3} \right ) < 0$
$t < 3$ and $(t - 3) < 0$

Note: $(t-2)$ and $(t-3)$ are squared in the denominator, making the denominator positive.

\[ R'(t) = -\frac {3 \left (t - \left (\frac {1 - 2 \sqrt {10}}{3} \right ) \right ) \left (t - \left (\frac {1 + 2 \sqrt {10}}{3} \right ) \right ) }{ (t-2)^2 (t-3)^2 } = neg \cdot \frac {pos \cdot neg \cdot neg}{pos \cdot pos} = neg \]

$R$ is decreasing.

On $\left ( \frac {1 - 2 \sqrt {10}}{3}, 2 \right )$,

$\frac {1 - 2 \sqrt {10}}{3}$ has an odd multiplicity, so the sign of $R'$ switches, $R'(t) > 0$.

Or,

$t > \frac {1 - 2 \sqrt {10}}{3}$ and $\left ( t - \frac {1 - 2 \sqrt {10}}{3} \right ) > 0$
$t < 2$ and $(t - 2) < 0$
$t < \frac {1 + 2 \sqrt {10}}{3}$ and $\left ( t - \frac {1 + 2 \sqrt {10}}{3} \right ) < 0$
$t < 3$ and $(t - 3) < 0$

Note: $(t-2)$ and $(t-3)$ are squared in the denominator, making the denominator positive.

$R$ is increasing.

On $\left ( 2, \frac {1 + 2 \sqrt {10}}{3} \right )$,

$2$ has an even multiplicity, so the sign of $R'$ does not switch sign, $R'(t) > 0$.

$R$ is increasing.

On $\left ( \frac {1 + 2 \sqrt {10}}{3}, 3 \right )$,

$\frac {1 + 2 \sqrt {10}}{3}$ has an odd multiplicity, so the sign of $R'$ switches sign, $R'(t) < 0$.

$R$ is decreasing.

On $(3, \infty )$,

$3$ has an even multiplicity, so the sign of $R'$ does not switch sign, $R'(t) < 0$.

$R$ is decreasing.

$\blacktriangleright $ Global (Extrema) Maximum and Minimum

From the behavior around the singularities, we know

\[ \lim \limits _{t \to 3^-}R(t) = -\infty \]

\[ \lim \limits _{t \to 3^+}R(t) = \infty \]

$R$ has no global maximum or minimum.

$\blacktriangleright $ Local (Extrema) Maximum and Minimum

We have two critical numbers.

For critical number $\frac {1 - 2 \sqrt {10}}{3}$:

$bullet$ On $\left ( -\infty , \frac {1 - 2 \sqrt {10}}{3} \right )$, $R$ is decreasing.

$bullet$ On $\left ( \frac {1 - 2 \sqrt {10}}{3} \right ), 2$, $R$ is increasing.

$R$ has a local minimum value at $\frac {1 - 2 \sqrt {10}}{3}$.

For critical number $\frac {1 + 2 \sqrt {10}}{3}$:

$bullet$ On $\left ( 2, \frac {1 + 2 \sqrt {10}}{3} \right )$, $R$ is increasing.

$bullet$ On $\left ( \frac {1 + 2 \sqrt {10}}{3} \right ), 3$, $R$ is decreasing.

$R$ has a local maximum value at $\frac {1 + 2 \sqrt {10}}{3}$.

$\blacktriangleright $ Range

The only thing we need to know is which is greater

\[ R \left ( \frac {1 - 2 \sqrt {10}}{3} \right ) \, \text { or } \, R \left ( \frac {1 + 2 \sqrt {10}}{3} \right ) \]

This will tell us if there is a gap in the range.
Hint: We can see a gap in the graph.

Instead of comparing these two function values directly, it might be easy to compare them to other numbers forcing a gap.

Like, we might show that

\[ R \left ( \frac {1 - 2 \sqrt {10}}{3} \right ) > -10 > R \left ( \frac {1 + 2 \sqrt {10}}{3} \right ) \]

\[ R\left ( \frac {1 - 2 \sqrt {10}}{3} \right ) = \frac {\left ( \frac {1 - 2 \sqrt {10}}{3} + 3 \right ) \left ( \frac {1 - 2 \sqrt {10}}{3} + 1 \right )}{\left ( \frac {1 - 2 \sqrt {10}}{3} - 2 \right ) \left ( \frac {1 - 2 \sqrt {10}}{3} - 3 \right )} \]

\[ = \frac {\left ( \frac {10 - 2 \sqrt {10}}{3} \right ) \left ( \frac {4 - 2 \sqrt {10}}{3} \right )}{\left ( \frac {-3 - 2 \sqrt {10}}{3} \right ) \left ( \frac {-8 - 2 \sqrt {10}}{3} \right )} \]

\[ = \frac { (10 - 2 \sqrt {10}) (4 - 2 \sqrt {10}) }{ (-3 - 2 \sqrt {10}) (-8 - 2 \sqrt {10}) } \]

\[ = \frac { 80 - 28 \sqrt {10} }{ 64 + 22\sqrt {10} } \]

\[ = \frac { 80 - 28 \sqrt {10} }{ 64 + 22\sqrt {10} } \cdot \frac { 64 - \sqrt {10} }{ 64 - 22\sqrt {10} } = \frac {11280 - 3552 \sqrt {10}}{-744} \]

Is this greater than $-10$ ?

\[ \frac {11280 - 3552 \sqrt {10}}{-744} > -10 \]

\[ 11280 - 3552 \sqrt {10} < -10 (-744) \]

\[ 11280 - 7440 < 3552 \sqrt {10} \]

\[ 3840 < 3552 \sqrt {10} \]

This is certainly true, since $3552 \cdot 2 = 7104 < 3552 \sqrt {10}$

Similarly, $-10 > R \left ( \frac {1 + 2 \sqrt {10}}{3} \right )$

The range has a gap.

The function has three connected components (from the singularities) and the range of each is

\[ \left [ R \left ( \frac {1 - 2 \sqrt {10}}{3} \right ), \infty \right ) \]

\[ \left ( -\infty , R \left ( \frac {1 + 2 \sqrt {10}}{3} \right ) \right ] \]

\[ (1, \infty ) \]

The range of $R$ is

\[ \left ( -\infty , R \left ( \frac {1 + 2 \sqrt {10}}{3} \right ) \right ] \cup \left [ R \left ( \frac {1 - 2 \sqrt {10}}{3} \right ), \infty \right ) \]

Everything agrees with the graph.

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2025-07-20 18:30:32