A Modeling Approach to Calculus: MA2C

The first step in a question about extreme values is to compute the rate of change \(f'(x)\text{.}\)

To apply the First Derivative Test, we need to complete sign analysis. Because \(f'(x)\) is defined and continuous everywhere, our critical points of \(f\) are the roots of \(f'(x)\text{.}\) We solve the equation by factoring:

There are two roots of \(f'\text{,}\) \(x=6\) and \(x=-2\text{,}\) which are the critical points of \(f\text{.}\)

Next, we perform sign analysis using the roots as the end points of the test intervals, which are \((-\infty,-2)\text{,}\) \((-2,6)\text{,}\) and \((6,\infty)\text{.}\) Using the factored function \(f'(x)=3(x+2)(x-6)\) makes it easier to find the signs without necessarily computing the final value. We can just look at the signs of each factor:

We now know \(f\) is increasing on \((-\infty,-2]\text{,}\) decreasing on \([-2,6]\text{,}\) and increasing on \([6,\infty)\text{.}\)

By the First Derivative Test, we now know that \(f\) has a local maximum at \(x=-2\) and a local minimum at \(x=6\text{.}\) The \(y\)-coordinate of the local maximum can be found using the formula for \(f(x)\text{:}\)

This is guaranteed to be the maximum value over the interval \((-\infty,6]\text{.}\) The \(y\)-coordinate of the local minimum can also be found:

which is guaranteed to be the minimum over the interval \([-2,\infty)\text{.}\) If we wanted to graph this function and show the local extrema, we would know that our window would need to include the \(x\)-values of \(x=-2\) and \(x=6\) as well these \(y\)-values.

In SageMath, we find the derivative formula using the diff command, which stands for the verb differentiate. The following script will define our function for SageMath and then ask it to show us the derivative where \(x\) is the independent variable.

We see that \(f(x)\) has a derivative

We can simplify this if we get a common denominator.

The denominator of \(f'(x)\) is never zero because \(x^2+3 \ge 3\) will never equal zero. So the sign can only change where \(3-x^2=0\) which occurs at two values, \(x=\pm\sqrt{3}\text{.}\) There are three intervals of interest to test, \((-\infty,-\sqrt{3})\text{,}\) \((-\sqrt{3},\sqrt{3})\text{,}\) and \((\sqrt{3},\infty)\text{.}\) We can find the signs of \(f'(x)\) using the values \(x=-2\text{,}\) \(x=0\) and \(x=2\text{.}\) The signs are summarized in the number line summary below.

We finish by interpreting our results.

• Because \(f'(x) \lt 0\) on \((-\infty,-\sqrt{3})\) and \(f'(x) \gt 0\) on \((-\sqrt{3},\sqrt{3})\text{,}\) we know \(f(x)\) has a local minimum at \(x=-\sqrt{3}\text{.}\) (Minimum over the interval \((-\infty,+\sqrt{3})\))
• Because \(f'(x) \gt 0\) on \((-\sqrt{3},\sqrt{3})\) and \(f'(x) \lt 0\) on \((\sqrt{3},\infty)\text{,}\) we know \(f(x)\) has a local maximum at \(x=\sqrt{3}\text{.}\) (Maximum over the interval \((-\sqrt{3},\infty)\))

A graph of the function is illustrated below.

This question considers finding global extrema for a function when it is restricted to different domains. Regardless of which domain we use, we will need to find the local extreme values. Local extreme values are identified from sign analysis of \(f'(x)\text{.}\) We find

Sign analysis begins by finding the roots, where \(f'(x)=0\text{.}\)

Because \(f'(x)\) is continuous, the roots determine the test intervals: \((-\infty,0)\text{,}\) \((0,\frac{8}{3})\text{,}\) and \((\frac{8}{3},\infty)\text{.}\) Testing one value of \(x\) from each interval in \(f'(x) = x(8-3x)\text{,}\) we find the signs summarized on the following number line.

We can interpret the sign analysis of \(f'(x)\) as characterizing the monotonicity of \(f(x)\text{.}\)

• \(f(x)\) is decreasing on \((-\infty,0)\text{.}\)
• \(f(x)\) is increasing on \((0,\frac{8}{3})\text{.}\)
• \(f(x)\) is decreasing on \((\frac{8}{3},\infty)\text{.}\)

Because \(f(x)\) is decreasing on the left of \(x=0\) and increasing on the right, \(f(x)\) has a local minimum at \(x=0\text{.}\) The value of the function at this minimum is

Similarly, because \(f(x)\) is increasing on the left of \(x=\frac{8}{3}\) and decreasing on the right, \(f(x)\) has a local maximum at \(x=\frac{8}{3}\text{.}\) The value of the function at this maximum is

From our sign analysis of \(f'(x)\text{,}\) we know that \(f(0)\) is the minimum value for the interval \((-\infty,\frac{8}{3}]\) and that \(f(\frac{8}{3})\) is the maximum value for the interval \([\frac{8}{3},\infty)\text{.}\) To complete the analysis of global extreme values, we need use limits to compare \(f(0)\) with points on the interval \((\frac{8}{3},\infty)\) and \(f(\frac{8}{3})\) with points on the interval \((-\infty,0)\text{.}\) To evaluate limits involving \(\pm \infty\text{,}\) we need to factor out the dominant power,

before using the limit arithmetic of infinity.

Let us now address the question of the global extreme values on each of the requested restricted domains.

1. Find the global extremes on the interval \((-\infty,\infty)\text{.}\)

   We have learned that on the interval \((-\infty,0)\text{,}\) \(f\) reaches all values between \(f(0)=0\) and \(\displaystyle \lim_{x \to -\infty}f(x)=\infty\text{.}\) Clearly, \(f(x)\) has no maximum value because it is unbounded above. Similarly, we learned that on the interval \((\frac{8}{3},\infty)\text{,}\) \(f(x)\) reaches all values between \(-\infty\) and \(f(\frac{8}{3})=\frac{256}{27}\) and is unbounded below. So \(f(x)\) has no minimum value. The range of \(f\) has been shown to be \((-\infty,\infty)\) and \(f\) has no global extreme values on \((-\infty,\infty)\text{.}\)

2. Find the global extremes on the interval \(D_1=[-1,3]\text{.}\)

   We know \(f\) is decreasing on \([-1,0)\text{,}\) a subset of \((-\infty,0)\text{,}\) so the maximum value on that interval is

   \begin{equation*} f(-1) = 4(-1)^2-(-1)^3 = 5\text{.} \end{equation*}

   We also know \(f\) is decreasing on \((\frac{8}{3},3]\) so the minimum value on that interval is

   \begin{equation*} f(3) = 4(3)^2 - (3)^3 = 9\text{.} \end{equation*}

   Comparing these to the local minimum \(f(0)=0\) and the local maximum \(f(\frac{8}{3})=\frac{256}{27}\text{,}\) we see that \(f(0)=0\) is the global minimum and \(f(\frac{8}{3})=\frac{256}{27}\) is the global maximum for the interval \(D_1=[-1,3]\text{.}\) When restricted to this domain, the range of \(f\) becomes \([0,\frac{256}{27}]\text{.}\)

3. Find the global extremes on the interval \(D_2=(-2,4]\text{.}\)

   We know \(f\) is decreasing on \((-2,0)\text{,}\) so that the left end-point provides an upper bound

   \begin{equation*} \lim_{x \to -2^+} f(x) = f(-2) = 4(-2)^2-(-2)^3=24\text{.} \end{equation*}

   This is not a maximum value because \(x=-2\) is not included in the domain. Because \(f\) is decreasing on \((\frac{8}{3},4]\) the minimum value on that interval is

   \begin{equation*} f(4) = 4(4)^2 - (4)^3 = 0\text{.} \end{equation*}

   The global minimum occurs in two locations,

   \begin{equation*} f(0) = f(4) = 0\text{.} \end{equation*}

   The value \(f(-2)=24\) is not a global maximum because \(x=-2\) is not included in the domain. However, \(f\) does include all values up to that value through the limit so that \(f\) has an upper bound of 24. When \(f\) is restricted to \(D_2=(-2,4]\text{,}\) the range is \([0,24)\text{.}\)

4. Find the global extremes on the interval \(D_3=[-2,2)\text{.}\)

   From the work above, we know \(f(-2)=24\) is the maximum value on \([-2,0)\text{.}\) Because the right end point \(x=2\) is to the left of the local maximum at \(x=\frac{8}{3}\text{,}\) we need to consider the interval of monotonicity \((0,2)\text{.}\) \(f\) is increasing on this interval and bounded above by the limit

   \begin{equation*} \lim_{x \to 2^-} f(x) = f(2) = 4(2)^2-2^3 = 8\text{.} \end{equation*}

   Our work shows that \(f\) has a global minimum at \(x=0\) with \(f(0)=0\) and a global maximum at \(x=-2\) with \(f(-2)=24\text{.}\) The range of \(f\) restricted to \(D_3=[-2,2)\) is \([0,24]\text{.}\)

The following figure illustrates the graph \(y=f(x)\) restricted to each domain. Be sure to compare the analysis that identified the global extremes with the graphs.

Our function of interest is defined by an accumulation

We can find local extrema in the same way as before. However, instead of solving an equation \(f'(x)=0\text{,}\) we can look at the graph to both find the roots and the signs of \(f'\text{.}\) The graph crosses the \(x\)-axis at \(x=-3\text{,}\) \(x=1\) and \(x=5\text{.}\) The signs of \(f'(x)\) are identified in the following sign analysis number line summary.

The Theorem 8.1.3 allows us to conclude that \(f(x)\) has a local maximum at \(x=-3\text{,}\) a local minimum at \(x=1\text{,}\) and another local maximum at \(x=4\text{.}\) It is possible to decide which maximum has a higher value by considering the signed area of the graph. In particular, because \(f'(x)\) has linear segments, we can compute the areas in question using elementary geometry to find

Using the splitting property of definite integrals, this implies

Consequently, \(f(4) = f(-3) + 1\) and \(f\) has a higher value at \(x=4\) than at \(x=-3\text{.}\)

We can find actual values if we use the initial value,

Because the integral goes right to left, we have \(dz \lt 0\) and the signed area will be negated. Again using geometry, we find

Using this point and the integrals above, we can quickly find

To find global extrema, we need to think about what happens to the left and right of these local extrema. The sign analysis of \(f'(x)\) shows that \(f\) is increasing on \((-\infty,-3)\text{.}\) Thus, \(f(-3)\) is the maximum value on \((-\infty,3]\text{.}\) The value of \(f\) is unbounded below on this interval,

We can see this by considering the integral \(\displaystyle \int_{-3}^{x}\) for \(x \lt -3\text{.}\) Because the integral goes right to left, \(dz \lt 0\text{,}\) and \(f'(z) \gt 0\) on this interval, the integral becomes more and more negative the further \(x\) goes to the left. With a similar argument using \(\displaystyle \int_4^x f'(z) \, dz\text{,}\) we find

and \(f\) is again unbounded below on the interval \((4,\infty)\text{.}\)

We conclude that \(f\) has a global maximum \(f(4) = 13.5\) and no global minimum. A graph of \(y=f(x)\) is shown below.