\(\blacktriangleright \) 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
with
\(R(t)\) is not a composition. It is a rational function.
Here is a quick DESMOS graph.
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.
\(R(2.5) = -77\)
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 \(\frac {pos \cdot pos}{neg \cdot neg} = positive\)
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 \(\frac {pos \cdot pos}{pos \cdot neg} = negative\)
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 \(\frac {pos \cdot pos}{pos \cdot neg} = negative\)
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 \(\frac {pos \cdot pos}{pos \cdot pos} = positive\)
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.
\(\blacktriangleright \) 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.
\(3 (3 t^2 - 2 t - 13) = 0\)
\(3 t^2 - 2 t - 13 = 0\)
Two critical numbers: \(\frac {1 - 2 \sqrt {10}}{3}\) and \(\frac {1 + 2 \sqrt {10}}{3}\)
Quick Check:
These agree with the graph, so we feel like our algebra is correct.
We can now factor \(R'(t)\)
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
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.
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.
\(R\) 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
\(R\) 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
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
Is this greater than \(-10\) ?
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
The range of \(R\) is
Everything agrees with the graph.
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More Examples of Analyzing Functions