Good Question 5 – 1998 AB2/BC2

Continuing my occasional series of some of my favorite teaching questions, today we look at the 1998 AP Calculus exam question 2. This question appeared on both the AB and BC exams. I use this problem to illustrate two very different questions that come up almost every time I lead a workshop or an AP Summer Institute. The first is if a limit is infinite, should you say “infinite” or “does not exist (DNE)”? The second is if the student solves the problems correctly, but by some other method, maybe even one not using the calculus, do they still earn full credit? In addition to discussing these two questions I’ll have a few suggestions for how to use this kind of question for teaching (maybe in other than a calculus class).

The question had the student examine the function f\left( x \right)=2x{{e}^{2x}} and, although it is easy enough to answer without, students were allowed to use their graphing calculator. A reasonable student probably looked at a graph of the function.

f\left( x \right)=2x{{e}^{2x}}

Part a: First the question ask student to explore the end behavior of the function by finding two limits: \underset{x\to -\,\infty }{\mathop{\lim }}\,f\left( x \right) and \underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right). The students should not depend on the graph here. As x\to -\infty , {{e}^{2x}}approaches zero and since the exponential function dominates the polynomial, \underset{x\to -\,\infty }{\mathop{\lim }}\,f\left( x \right)=0. In passing note that for x < 0 the function is negative and approaches zero from below. No work or explanation was required, but when teaching things like this be sure students know and can explain their answer without reference to their calculator graph.  For the second limit, since both factors increase without bound \underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)=\infty  If the student wrote \underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)=\text{DNE}, he received full credit.

Infinity is not a number, so there really is no limit in the second case; the limit DNE. But there are other ways a limit may not exist such as a jump discontinuity or an oscillating discontinuity.  DNE covers these as well as infinite limits. Saying a limit is infinite tells us more about the limit than DNE. It tells us that the function increases without bound; that eventually it becomes greater than any number.

Both answers are correct.

But we’re not done with this yet. We will come back to it before the question is done.

Part b: Students were asked to find and justify the minimum value of the function. Using the first derivative test, students proceeded by finding where the derivative is zero..

{f}'\left( x \right)=\left( 2x \right)\left( 2{{e}^{2x}} \right)+2{{e}^{2x}}=2{{e}^{2x}}\left( 2x+1 \right)=0

x=-\frac{1}{2}

f\left( -\tfrac{1}{2} \right)=2\left( -\tfrac{1}{2} \right){{e}^{2\left( -\tfrac{1}{2} \right)}}=-\frac{1}{e}\approx 0.368\text{ or }0.367

Justification: If x<-\tfrac{1}{2},\ {f}'\left( x \right)<0 and if x>-\tfrac{1}{2},\ {f}'\left( x \right)>0, therefore the absolute minimum is -\frac{1}{e} and occurs at x=-\frac{1}{2}.

All pretty straightforward

Part c: This part asked for the range of the function. Here the student must show that if he wrote DNE in part a, he knows that in fact the function grows without bound.

Putting together the answers from part a and part c, the range is f\left( x \right)\ge -\frac{1}{e}, which may also be written as \left[ -1/e,\infty\right). (The decimals could also be used here.)

Part d: asked students to consider functions given by y=bx{{e}^{bx}} where b was a non-zero number. The question required students to show that the absolute minimum value of all these functions was the same.

Most students did what was expected and preceded as in part b. The work is exactly the same as above except that all of the 2s become bs. The absolute minimum occurs at x=-\frac{1}{b} and y\left( -\tfrac{1}{b} \right)=-\frac{1}{e}.

BUT ….

Other students found a way completely without “calculus.” Can you find do that?

They realized that the given function as a horizontal expansion or compression, possibly including a reflection over the y-axis, of and therefore the range is the same for all these functions and so the minimum value must be the same. This received full credit. The rule of thumb is “don’t take off for good mathematics.”

Pretty cool!

The graphs of several cases are shown below

y=bx{{e}^{bx}}
b = -5 in blue, b = -1 in red, b = 2 in green, and b = 4 in magenta.

 

Teaching Suggestions

I can see using this in a pre-calculus class. The calculus (finding the minimum for b = 2 or in general) is straightforward. In a pre-calculus setting as an example of transformations it may be more useful. You could give students 6, or 8, or 10 examples with different values of b, both positive and negative.

  1. First ask students to investigate the end behavior by finding the limits as x approaches positive and negative infinity. The results will be similar. Have them write a summary considering two cases: b > 0 and b < 0.
  2. Graphing calculators have built-in operations that will find the x-coordinates or both coordinates of the minimum point of a function. Since we’re concerned with the transformation and not the calculus, let students use their graphing calculators to find the coordinates of the minimum point of each graph (as decimals). See if they can determine the x-coordinate in terms of b. They should also notice that y-coordinates will all be the same (about -0.367880).
  3. Finally, set the class to proving using their knowledge of transformation that the minimums are really all the same.
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