Stars and functions

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Two young mathematicians discuss stars and functions.

Check out this dialogue between two calculus students (based on a true story):

Devyn: Riley, did you know I like looking at the stars at night?
Riley: Stars are freaking awesome balls of nuclear fire whose light took thousands of years to reach us.
Devyn: I know! But did you know that the best way to see a very dim star is to look near it but not exactly at it? It’s because then you can use the “rods” in your eye, which work better in low light than the “cones” in your eyes.
Riley: That’s amazing! Hey, that reminds me of when we were talking about the two functions
\[ f(x) = \frac {x^2-3x+2}{x-2}\qquad \text {and}\qquad g(x)= x-1, \]
which we now know are completely different functions.
Devyn: Whoa. How are you seeing a connection here?
Riley: If we want to understand what is happening with the function
\[ f(x) = \frac {x^2-3x+2}{x-2}, \]
at $x=2$, we can’t do it by setting $x=2$. Instead we need to look near $x=2$ but not exactly at $x=2$.
Devyn: Ah ha! Because if we are not exactly at $x=2$, then
\[ \frac {x^2-3x+2}{x-2} = x-1. \]

Let $f(x) = \frac {x^2-3x+2}{x-2}$ and $g(x) = x-1$. Which of the following is true?

$f(x) = g(x)$ for every value of $x$. There is no $x$-value where $f(x) = g(x)$. $f(x) = g(x)$ when $x\ne 2$.

When you evaluate

\[ f(x) = \frac {x^2-3x+2}{x-2}, \]

at $x$-values approaching (but not equal to) $2$, what happens to the value of $f(x)$? The value of $f(x)$ approaches $\answer {1}$.

Just from checking some values, can you be absolutely certain that your answer to the previous problem is correct?

yes no

Here you only have information about a few specific points on the graph. There are infinitely many $x$-values close to, but not equal to, $x = 2$. Hence we cannot be completely certain.

2025-01-06 19:55:05