A Modeling Approach to Calculus: MA2C

Because the denominator of \(f(x)\) is never zero, \(x^2+1 \ne 0\text{,}\) \(f(x)\) is continuous everywhere. Consequently, the Extreme Value Theorem guarantees that \(f\) will have a maximum and minimum value on the closed interval \([-2,2]\text{.}\)

First, we compute \(f'(x)\text{,}\) which involves the quotient rule.

Next, we find critical points. Again, the denominator is nonzero, so \(f'(x)\) will be defined and continuous for every value \(x\text{.}\) We only need to find solutions to \(f'(x)=0\text{,}\) which are solutions to \(-x^2+2x+1=0\text{.}\) The quadratic formula gives

The critical values are \(x=1-\sqrt{2}\approx 0.414\) and \(1+\sqrt{2} \approx 2.414\text{.}\)

If we were to test these critical values as turning points, we would look at the signs of \(f'(x)\) on the intervals formed by the critical points. The sign analysis is summarized with the number line below, showing that \(x=1-\sqrt{2}\) is a local minimum and \(x=1+\sqrt{2}\) is a local maximum.

To find the extreme values on the interval, we really just need to compare the values of \(f(x)\) at the end points of the interval with the critical points that are inside the interval. Because \(x=1+\sqrt{2}\) is outside \([-2,2]\text{,}\) we do not include that point.

We finish by interpreting our results. The maximum value of \(f\) on the interval \([-2,2]\) is \(\frac{1}{5}\text{,}\) occurring at \(x=2\text{.}\) The minimum value of \(f\) on the interval is \(\frac{\sqrt{2}}{4-2\sqrt{2}} \approx -1.207\text{,}\) occurring at \(x=1-\sqrt{2} \approx 0.414\text{.}\) A graph of the function showing these extremes is given below.

The function is continuous. If the interval was closed, \([-2,2]\text{,}\) the Extreme Value Theorem would guarantee that it had a maximum and minimum value. By excluding the end points, we might no longer achieve one or both of those values.

First, find \(f'(x)=3x^2+2x-2\text{.}\) Use this to find critical points, where \(f'(x)=0\text{.}\) The quadratic formula is needed:

The critical values are \(x=\frac{-1-\sqrt{7}}{3} \approx -1.215\) and \(x=\frac{-1+\sqrt{7}}{3} \approx 0.5486\text{.}\) Both critical values are in the interval.

We now compare the values of the function at the critical values and at the end points.

The minimum value for \(f\) on the interval \([-2,2]\) is approximately 1.369 at \(x=\frac{-1+\sqrt{7}}{3}\text{;}\) the maximum value is 10 at \(x=2\text{.}\) However, when working with the open interval \((-2,2)\text{,}\) the maximum value is a limit value that is not achieved. So \(f\) does not have a maximum value on \((-2,2)\text{,}\) though it does have the same minimum value.

The function \(f(x)=\frac{10x-15}{x^2}\) has a discontinuity at \(x=0\) because division by zero is undefined. This corresponds to a vertical asymptote at \(x=0\text{.}\) Because the question does not give an interval, we must be considering the entire domain, \((-\infty,0) \cup (0,\infty)\text{.}\) Extreme values are not guaranteed.

We begin by finding critical values. The derivative requires the quotient rule.

The critical value is the solution to \(-10x+30=0\text{,}\) or \(x=3\text{.}\) When doing sign analysis, we also need to use the discontinuity \(x=0\) to create the intervals that are tested.

The sign analysis on \(f'(x)\) informs us about the vertical asymptote. The function decreases immediately to the left of \(x=0\text{,}\) letting us know

Similarly, the function is increasing immediately to the right of \(x=0\text{,}\) showing that

This is enough to guarantee that \(f(x)\) does not have a minimum value because it is unbounded in the negative direction.

The critical point \(x=3\) is a turning point corresponding to a local maximum. The value of the function is \(f(3)=\frac{15}{9}=\frac{5}{3}\text{.}\) To see if this is a global maximum, we need to compare it with the ends of the intervals, as \(x \to \pm \infty\text{.}\) These limits are both zero.

These limits show that \(y=0\) is a horizontal asymptote for \(f(x)\) as \(x \to \pm \infty\text{.}\)

Interpreting our results, we see that \(f(x)\) has a maximum value of \(\frac{5}{3}\) at \(x=3\) and no minimum value due to the infinite limit at \(x=0\text{.}\)

Physically, the driving frequency must be a non-negative value, so \(\omega \ge 0\text{.}\) So we will look for extreme values on the domain \([0,\infty)\text{.}\) The denominator involves the sum of two squares which can not be simultaneously equal to zero. Consequently, \(A(\omega)\) is a continuous function defined for all values of \(\omega\text{.}\) We need to find the critical values by solving \(A'(\omega)=0\text{.}\) Computing \(A'\) involves repeated use of the chain rule.

Critical values are solutions to the equation \(2 \omega (\omega^2-\omega_0^2+2\alpha^2) = 0\text{.}\) Solutions occur when \(\omega=0\) and \(\omega^2 = \omega_0^2-2\alpha^2\text{,}\) which only occurs when \(\omega_0^2 \gt 2 \alpha^2\text{.}\)

When \(\omega_0^2 \gt 2 \alpha^2\text{,}\) the sign of \(A'(\omega)\) changes sign at the critical value at

since the factor \(\omega^2 - (\omega_0^2-2\alpha^2)\) changes sign. The denominator will always be positive. This allows us to determine the sign analysis of \(A'(\omega)\text{.}\)

The sign analysis of \(A'(\omega)\) shows that \(A(\omega)\) is increasing for \(\omega\) in the interval \([0,\omega^*]\) and decreasing for \(\omega\) in the interval \([\omega^*,\infty)\text{.}\) Consequently, \(A\) achieves a maximum value when \(\omega = \omega^*\text{.}\) That maximum value is

A typical resonance response curve is shown below for this case (\(\omega_0=1\) and \(\alpha=0.25\)).

In the case that \(\omega_0^2 \lt 2 \alpha^2\) and the only critical value is \(\omega=0\text{,}\) the factor in the numerator of \(A'(\omega)\) given by \(\omega^2 - \omega_0^2 + 2\alpha^2\) will always be positive. This implies that \(A'(\omega) \lt 0\) for all \(\omega \gt 0\text{,}\) meaning that \(A(\omega)\) is a decreasing function whose maximum must be at the end-point \(\omega=0\text{.}\)

We start by looking at the graph of the data. Let \(P\) be the population size and let \(b\) be the per capita birth rate. We are interested in the function \(P \mapsto b\text{,}\) so the graph plots points \((P, b)\text{.}\)

The relationship between these points is decreasing. A linear model looks like it could be appropriate. Using linear regression on our data, we find a model

The total birth rate is defined as \(B(P) = b(P) \cdot P\text{,}\) which according to our model is given by

This is our objective function, the quantity we want as large as possible.

To find the maximum value of \(B\text{,}\) we compute \(B' = 0.019878 - 3.2846 \times 10^{-5} P\) and solve \(B'=0\) to find the critical point.

Because \(B'' = -3.2846 \times 10^{-5}\) is negative, we know that \(B\) is a concave down function and the critical value corresponds to a maximum value. The maximum total birth rate is

occurring for \(P=605.2\text{.}\) A population should be an integer value, so we would expect the maximum birth rate to occur when \(P=605\text{.}\)

However, our prediction is dependent on the model that we chose. What if we used a different equation for our original per capita birth rate? The original data points have the appearance of being concave up. So maybe we should use an exponential regression curve. This gives

The graph below shows the two models for per capita birth rate with the data.

Using the exponential model for the per capita birth rate, we obtain a modified objective function

The derivative is

The critical value must solve \(1-0.001125P\) since the exponential factor is always positive. This gives \(P = \frac{1}{0.001125} \approx 888.9\text{.}\) Sign analysis of \(B'(P)\) shows that this critical value corresponds to a maximum value, \(B(888.9)=6.763\text{.}\)

Notice how two different models can yield significantly different predictions. The two models are illustrated together in the figure below. Notice that the original, quadratic model has the non-physical predication of a negative number of births for sufficiently large population sizes.