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Effective GDC Use #3

Saturday 17 November 2018

Concerning effective GDC use, I try to avoid giving my students the impression that mindlessly pushing buttons on a GDC will allow them to successfully answer an exam question. Although there may be exam questions where this occurs (more on this in future blog posts), I endeavour to give my students questions which combine smart use of their GDC with some clear thinking.

Early in Maths SL or HL, students will be expected to identify the domain and range of a given function. Although not inherently difficult, it’s possible to compose questions about the domain and range of a function in such a way that wise use of a GDC is necessary but not sufficient to obtain the correct answers. A student will also need to apply some number sense and knowledge about certain functions. One of the habits I try to engrain upon my students with such questions is to make an educated guess about the domain and range of a function before graphing the function on a GDC, and then make use of a GDC to confirm / modify the determination of the function’s domain and range. Although not really exam-like, I like these kind of questions because they can be used early in either Maths SL or HL to promote thoughtful application of a GDC. Here are two examples.

Question:  Determine the domain and range of each function.

(a)  $$f\left( x \right) = \frac{1}{{{x^2} + 3x - 10}}$$          (b)  $$g\left( x \right) = \sqrt {\frac{{8x - 4}}{{x - 3}}}$$

Solution:

(a)  A student should immediately factorise the denominator to identify the vertical asymptotes for the graph of the function.

$$f\left( x \right) = \frac{1}{{{x^2} + 3x - 10}} = \frac{1}{{\left( {x + 5} \right)\left( {x - 2} \right)}}$$

Thus, the graph of f has vertical asymptotes (and discontinuities) at $$x = - 5$$ and at $$x = 2$$.  The domain is certain to be $$x \in \mathbb{R},\;x \ne - 5,\;x \ne 2$$.  Determining the range is much less clear.  One can figure out that the graph of f will have a horizontal asymptote $$y = 0$$ (x-axis) because as x becomes either a very large positive or very large negative number the $${x^2}$$ term will dominate and cause the denominator $${x^2} + 3x - 10$$ to become a large positive number.  Thus, as $$x \to + \,\infty$$ or as $$x \to - \,\infty$$ the value of $$\frac{1}{{{x^2} + 3x - 10}}$$ approaches zero from positive values meaning that the graph of f will be asymptotic to the x-axis from above (i.e. the end behaviour of f).  So, an educated guess is that the range is $$y \in \mathbb{R},\;y \ne 0$$.

Graphing on a GDC (see images below) confirms the domain as stated, but the ‘gap’ in y values is more than just zero.  The downward-opening branch of the graph of f appears to have a maximum value less than zero.  Analyzing the graph on the GDC reveals that this downward-opening branch of the graph has a maximum at $$\left( { - 1.5, - \,0.08163} \right)$$.  It’s clear that the x-coordinate of this local maximum point is exactly $$- 1.5$$ but what about the y-coordinate?  Finding it exactly requires some effort.  Somewhat surprisingly, it’s more of an effort on the TI-Nspire than the TI-84.  After converting to a fraction, it’s revealed that the exact value of the local maximum’s y-coordinate is $$- \frac{4}{{49}}$$.  Therefore, the range of f is $$y \le - \frac{4}{{49}},\;y > 0$$Note:  Although $$- \frac{4}{{49}} \approx - \,0.0816$$, it is incorrect to state the range as $$y \le - 0.0816,\;y > 0$$ or as $$y < - 0.0816,\;y > 0$$.  The ‘endpoint’ of an inequality used for indicating the domain or range must be expressed exactly.  For example, the value $$y = - \,0.8162$$ is less than $$- \,0.816$$; but $$- \frac{4}{{49}} \approx - \,0.081632653061...$$ and $$- \,0.8162$$ is not less than $$- \,0.081632653061...$$.  Hence, $$y = - \,0.8162$$ is not in the range but expressing the range as either $$y \le - 0.0816,\;y > 0$$ or $$y < - 0.0816,\;y > 0$$ would incorrectly indicate that $$y = - \,0.8162$$ is in the range.

(b)   For determining the domain of g we know that the graph of $$g\left( x \right) = \sqrt {\frac{{8x - 4}}{{x - 3}}}$$ has a vertical asymptote (and discontinuity) at $$x = 3$$.  It is also necessary to find where $$\frac{{8x - 4}}{{x - 3}} > 0$$ which takes a bit of effort (remember this is for the purpose of making an ‘educated guess’ before looking at a graph of the function on a GDC).  A sign chart will work for solving the inequality. So, it seems that the domain should be $$x < \frac{1}{2},\;x > 3$$.

Working out the range initially seems less tricky than the previous example.  Taking the square root can only give non-negative results.  Hence, an educated guess for the range is $$y \ge 0$$. A graph of g confirms the conjecture for the domain but appears to indicate there is a horizontal asymptote with an equation somewhere around $$y = 3$$. As with the function in part (a), considering the end behaviour of the function will lead to an exact equation for the horizontal asymptote.  When $$x \to + \,\infty$$ or $$x \to - \,\infty$$, the x terms in the numerator and denominator of $$\frac{{8x - 4}}{{x - 3}}$$ will dominate so that the value of $$\frac{{8x - 4}}{{x - 3}}$$ will come ever closer to $$\frac{8}{1}$$.  Thus, the equation of the horizontal asymptote is $$y = \sqrt 8 = 2\sqrt 2$$; also indicating where there will be a ‘gap’ in the range.  Therefore, the correct range for function g is $$0 \le y < 2\sqrt 2 ,\;y > 2\sqrt 2$$.