Far Out!

A monster problem for Halloween.

A while ago I suggested you look at $\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln \left( {{x}^{5}} \right)}{{{x}^{0.02}}}$ , which using the dominance idea is zero. Of course your students may try graphing or a table. Here’s the graph done by a TI-Nspire CAS. Note the scales.

This is not the way to go. Since the function is increasing near the origin, but the limit at infinity is zero there must be a maximum point where the function starts decreasing. And as the expression can never be negative once x > 1, there must be a point of inflection where the graph becomes concave up and can thereafter approach the x-axis from above as a horizontal asymptote. The maximum can be found by hand which makes for some great algebra manipulation practice:

$\displaystyle \frac{d}{dx}\left( \frac{\ln \left( {{x}^{5}} \right)}{{{x}^{0.02}}} \right)=\frac{{{x}^{0.02}}\tfrac{5{{x}^{4}}}{{{x}^{5}}}-\ln \left( {{x}^{5}} \right)\left( 0.02{{x}^{-0.98}} \right)}{{{x}^{0.04}}}$

$\displaystyle \frac{d}{dx}\left( \frac{\ln \left( {{x}^{5}} \right)}{{{x}^{0.02}}} \right)=\frac{{{x}^{-0.98}}\left( 5-\left( 0.10 \right)\ln \left( x \right) \right)}{{{x}^{0.04}}}=\frac{50-\ln \left( x \right)}{10{{x}^{1.02}}}$

Setting this equal to zero and solving gives $x={{e}^{50}}\approx 5.185\times {{10}^{21}}$

The second derivative is $\displaystyle \frac{{{d}^{2}}}{d{{x}^{2}}}\left( \frac{\ln \left( {{x}^{5}} \right)}{{{x}^{0.02}}} \right)=\frac{-510+10.2\ln \left( x \right)}{100{{x}^{2.02}}}$

and is zero when x = $\displaystyle {{e}^{\frac{520}{10.2}}}\approx 1.382\times {{10}^{22}}$

Okay, I skipped a few steps here, but you can challenge your students with that. Since we’re really interested in the solution here more than the solving ,this is really a good place to use a CAS calculator.

The first line in the figure above defines the function to save typing it each time. The second line finds the x-coordinate of the maximum point (how do we know this is a maximum?) and the third finds the x-coordinate of the point of inflection. Much simpler this way!

Take a minute to consider the numbers. They are BIG! In fact, if the units on our graph paper are centimeters, then the maximum point is a little over 5,480 light-years away from the origin! The point of inflection is about 2.665 times farther at more than 14,607 light-years away!

Meanwhile the maximum value is only 91.9699 cm. That’s right centimeters, less than a meter. And the y-coordinate of the point of inflection is about 91.9524 cm. A drop of 0.0175 cm. in a horizontal distance of a little over 9,127 light-years.

Some problems are a lot less scary if done with technology.

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Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Far Out!

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