Domain: Neither the exponential or the polynomial pieces of this function have any restrictions, so the domain is all real numbers.
Range: The range it hard to determine at the start. I will return to this.
Continuity: This is an elementary function, so it is continuous on its domain. The domain is all real numbers, so this function is continuous everywhere.
Intercepts: To find the \(y\) intercept, I simply evaluate the function at 0.
There is a \(y\) intercept at \((0,-6)\text{.}\) To find the \(x\) intercept, I need to solve \(f(x) = 0\text{.}\) Note that the exponential term can never be zero, so only the polynomial term can produce zero. I use the quadratic formula, since this quadratic doesn’t have whole number roots.
We have \(x\) intercepts at two points: \(x = \frac{-3
+ \sqrt{33}}{2} \doteq 1.372\) and \(x = \frac{-3 -
\sqrt{33}}{2} \doteq -4.372\text{.}\)
Symmetry: There is no symmetry here. There is no periodic symmetry because this is not a trigononmetric function. There is no even or odd symmetry because \(f(-x)\) does not simplify to either \(f(x)\) or \(-f(x)\text{.}\)
Limits and Asymptotoes: There are no breaks in the domain, so there is no possibility of vertical asymptotes. We can look at the limits to \(\pm \infty\) for horizontal asymptotes.
Asymptotically, the exponential decay is dominant over the polynomial growth, so the function approaches 0. This means that the function has a horizontal asymptote \(y=0\) in the positive direction.
For negative numbers, both the polynomial and exponential functions grow without bound. There is no horizontal asymptote in the negative direction.
First Derivative: We take the derivative. For ease of working with is for the future steps, I’ll factor out the exponential term and collect the polynomial terms after taking the derivative.
I set the derivative equal to zero and solve for the critical points. Notes, again, that the exponential can never be zero, so I only need to look at the quadratic.
I have two critical points. Their approximate values ate \(x \doteq -3.541\) and \(x \doteq 2.251\text{.}\) I move on to the intervals determined by these critical points. The domain is all reals, so there are three intervals. Note also that the exponential term is always positive, so the sign of the derivative is entirely determined by the quadratic.
This gives me a minimum at the first critical point and a maximum at the second. With the extrema and the intervals of increase and decrease, I am done with the first derivative.
Second Derivative: I differentiate again. For ease of working with is for the future steps, I’ll again factor out the exponential term and collect the polynomial terms after taking the derivative.
I set the second derivative equal to zero and solve for the potential inflection points. Notes, again, that the exponential can never be zero, so I only need to look at the quadratic.
I have two potential infelction points. Their approximate values ate \(x \doteq -2.702\) and \(x \doteq 3.702\text{.}\) I move on to the intervals. The domain is all reals, so I have three intervals. Note also that the exponential term is always positive, so the sign of the second derivative is entirely determined by the quadratic.
Range again: I know the function diverges to infinity as \(x \rightarrow -\infty\text{,}\) so there is no upper bound. However, the minimum is a lower bound, since the function has a horizontal asymptote in the positive direction. The minimum happens at \(x = \frac{-1-\sqrt{37}}{2}\text{.}\) Evaluating the function at this point is awkward, but I asked a computer for an approximate value: \(f \left(
\frac{-1-\sqrt{37}}{2} \right) \doteq -140.915\text{.}\) Therefore, the range (in approximate values) if \((-140.914, \infty)\)
Figure 11.3.1 shows the graph of the function, matching out analysis above.
Domain: To avoid dividing by zero, I need to exclude \(x =
\frac{5}{4}\text{.}\) The domain is all real numbers except \(x = \frac{5}{4}\)
Range: The range it hard to determine at the start. I will return to this.
Continuity: This is an elementary function, so it is continuous on its domain. There are two pieces of continuity: to the right and to the left of the undefined point \(x = \frac{5}{4}\text{.}\)
Intercepts: To find the \(y\) intercept, I simply evaluate the function at 0.
There is a \(y\) intercept at \(\left(0,\frac{9}{5}
\right)\text{.}\) To find the \(x\) intercept, I need to solve \(f(x) = 0\text{.}\) To do this, I only need to look at the numerator, since a rational function is 0 only when its numerator is zero.
\begin{align*}
f(x) \amp = \frac{x-9}{4x-5} = 0 \\
x - 9 \amp = 0 \\
x \amp = 9
\end{align*}
The point \(x=9\) is in the domain, so \((9,0)\) is an intercept.
Symmetry: There is no symmetry here. There is no periodic symmetry because this is not a trigononmetric function. There is no even or odd symmetry because \(f(-x)\) does not simplify to either \(f(x)\) or \(-f(x)\text{.}\)
Limits and Asymptotoes: There is an undefined point at \(x =
\frac{5}{4}\text{,}\) so I take the limits approaching this from the left and the right to see if there is a vertical asymptote. (I do the two sides seperately to give me more information about whether to draw the function approaching positive or negative \(\infty\text{.}\)0
The numerator approaches a finite number and the denominator approaches 0, so this does diverge to \(\pm \infty\text{.}\) Looking at the sign, for numbers slightly larger than \(\frac{5}{4}\text{,}\) the numerator is negative and the denominator is positive, so this limit approahces \(-
\infty\text{.}\)
The numerator approaches a finite number and the denominator approaches 0, so this does diverge to \(\pm \infty\text{.}\) Looking at the sign, for numbers slightly smaller than \(\frac{5}{4}\text{,}\) the numerator is negative and the denominator is also negative, so this limit approahces \(\infty\text{.}\) This means that the line \(x = \frac{5}{4}\) is a vertical asymptote.
I can look at the limits to \(\pm \infty\) for horizontal asymptotes.
The same behaviour holds in the negative direction as well. Both numerator and denominator will be negative, which cancels out, leaving just the ratio of the leading coefficients.
First Derivative: I take the derivative using the quotient rule.
I set the derivative equal to zero and solve for the critical points. However, this fraction has a constant numerator, so it can never be zero. There are no critical points, hence no possible minima or maxima. However, I still have to look at the intervals to see where the function is increasing or decreasing. The domain has two intervals, since it is divided up by the undefined point \(x = \frac{5}{4}\text{.}\)
I set the second derivative equal to zero and solve for the potential inflection points. Again, however, there are no solutions, so there are no inflection points. I still analyze the intervals to see where the function is concave up and concave down. The intervals are the same as for the first derivative, seperated by the undefined point.
Range again: The function has a horizontal asymptote at \(y
= \frac{1}{4}\text{.}\) To the left of \(x = \frac{5}{4}\text{,}\) the function is growing up from this asympote, approaching \(\infty\) at the verical asymptote. Therefore, the range includes all numbers greater than \(\frac{1}{4}\text{.}\) On the right of \(x = \frac{5}{4}\text{,}\) the function starts with a vertical asymptote and grows up to the horizontal asymptote at \(y = \frac{1}{4}\text{.}\) Therefore, all number less than \(\frac{1}{4}\) are also in the range. I conclude that the range is all real numbers except \(y =
\frac{1}{4}\text{.}\)Figure 11.3.2 shows the graph of the function, matching out analysis above.
Figure11.3.2.Second Curve Sketching Activity
Subsection11.3.2Model Analysis
Activity11.3.3.
Analyze and critique a model of population (in millions) depending on time (in years) give by the function
The asymptotic order of the numerator and denominator are the same, so the limit is the ratio of the leading coefficients. In the long term, this population approaches 4 million.
Derivative: I take the derivative to determine intervals of increase and decrease.
Assuming that I restrict to the domain \(t \gt
0\) (a reasonable assumption for most models which depend on time), this derivative is always positive. Therefore, the population is always increasing. In the limit, the derivative is getting very small, so the rate of increase is getting smaller and smaller over time as the population approaches 4 million.
Holistic analysis: These three observations give me a reasonable understanding of this model. This is like logistic growth (though not logistic growth, since it doesn’t conform to the specific pattern with exponential functions). It starts with modest growth which slows over time as it approaches a steady value. This seams like an entirely reasonable model of population.
Activity11.3.4.
Analyze and critique a model of temperature (in degree celcius) depending on ocean depth (in meters) given by the function
Starting value: I evaluate at \(d=0\) to get \(T(0) =
16 - 0 = 16\text{.}\) The temperature as the surface of the water is 16 degree celcius.
Derivative: I take the derivative to determine the increase or decrease of the model. (With the variable \(d\text{,}\) Leibniz notation is somewhat awkward, so I’ll use Newton’s notation for this derivative).
This derivative is always negative (assume the domain \(d
\gt 0\text{,}\) since a negative depth doesn’t make sense), so the temperature is always decreasing.
Long term behaviour: I take the limit as \(d \rightarrow
\infty\text{.}\)
The temperature gets lower and lower without bound.
Holistic analysis: The starting temperature seems reasonable: some oceans certainly have a surface water temperature of 16 degrees. The fact that the temperature decreases as it descend also makes sense: I expect the water to be cooler as depth increases. For example, if \(d =
900m\text{,}\) then \(T(900) = 16 - 10 = 6\text{.}\) At 900 meters below the surface, the temperature has slowly dropped to 6 degree, which is believable.
However, the long term behaviour past this point because more problematic. Eventually, there is a negative value, which isn’t that likely for deep oceans. If I allow the depth to continue further, I could get very large negative numbers, which make no sense. I can conclude this might be a reasonable model for the first while, maybe a kilometer or two of depth, but eventually the model breaks down and is no longer useful.
Activity11.3.5.
Analyze and critique a model of tree height (in cm) depending on time (in years) given by the function
The term \(3 \pi \cos t\) has minimum value \(-3\pi
\doteq 9.42\text{.}\) When I add 12, even with this minimum value, the derivative is always positive. The derivative is relatively small at these points in the period of cosine, but larger at other points in the period.
Holistic analysis: The tree starts at 4cm. This is reasonable if I started measuring a seedling. The growth rate is always possible, which is also good -- trees don’t normally start strinking. So far, this seems good.
Can I explain the sine term? Why would there be a oscillation in the growth? The period of this sine term is 1, and our units are years, so the period is 1 year. Oscillation of a growth rate in a year could be explained by seasonon variation, with (assuming Northern hemisphere) slower growth in the winter and faster growth in the summer. In this is the case, the overall rate of growth would be 12 cm per year (which is reasonable for some tree species), but less that 3 cm per year in the slow season and closer to 20 cm per year in the fast season. This makes sense.
However, eventually this model becomes unreasonable. Linear growth is not sustainable for ever; eventually the tree will approach its maximum height and growth will slow. This models looks like a reasonable choice for the first number of years, but eventually must be discarded as the tree approaches maturity.
Activity11.3.6.
Analyze and critique a model of population (in thousands) depending on time (in years) given by the differential equation
\begin{equation*}
\frac{dp}{dt} = 10 p (p-40).
\end{equation*}
Solution.
Here I only have a differential equation, not a function. I could try to solve the differential equation (though I haven’t covered such equations in this course), but I could also use implicit methods.
The phase line analysis (not drawn here) shows that there is a steady state at \(p=40\text{.}\) The trajectory below 40 is downward and the trajectory above 40 is upwards. Therefore, 40 is an unstable steady state. If the population starts below 40, it will decay to zero. If the polulation start above 40, it will grow without bound.
This seems mostly reasonable. For whatever reasons, 40 thousand is the minimum sustainable level for this population. The only limitation of this model is likely its long term behaviour, since growth without bound is not sustainable forever.
Activity11.3.7.
Analyze and critique a model of radioactive contamination (in grays) depending on time (in years) given by the function
Continuity: This is a piecewise model, so the first question is a question of continuity. We’ll take the limits of the functions approaching the crossover points.
At both the crossover points, the limit from the right and the limit from the left are the same. The model is continuous.
Behaviours: there are three behaviours. For the first 100 years, there is exponential decay. Then, for another 100 years, there is sinusoidal oscillation, with a relatively small amplitude. Finally it reverts to exponential decay.
Long term behaviour: the limit (using the third function) as \(t \rightarrow \infty\) is 0, since this is an exponential decay function. In the long term, the radiation dissipates away.
Holistic analysis: exponential decay is expected behaviour for radiation, so that two exponential pieces are reasonable. The middle piece needs a bit more explanation: instead of decay, there is an oscillation. This is a bit odd. It may be that there is some kind of periodic source of new radiation: waste being produced by some industrial process, perhaps, or radiative volcanic material from a natural source perhaps. In any case, something happens at 100 years to start this new behaviour and then something else happens at 200 years to stop this new behaviour and return to exponential decay. This may be plausible.
Subsection11.3.3Conceptual Review Questions
How do you connect many of the tools in the course together to get a holistic idea of the shape and behaviour of a function?
What is the qualitative analysis of functions? What does it mean to discuss the behaviour of a function?
