yes no
Two young mathematicians examine one (or two!) functions.
Check out this dialogue between two calculus students (based on a true story):
- Devyn
- Riley, I have a pressing question.
- Riley
- Tell me. Tell me everything.
- Devyn
- Think about the function \[ f(x) = \frac {x^2 - 3x + 2}{x-2}. \]
- Riley
- OK.
- Devyn
- Is this function equal to \(g(x) = x-1\)?
- Riley
- Well if I plot them with my calculator, they look the same.
- Devyn
- I know!
- Riley
- And I suppose if I write \begin{align*} f(x) &= \frac {x^2 - 3x + 2}{x-2} \\ &= \frac {(x-1)(x-2)}{x-2} \\ &= x-1 \\ &= g(x). \end{align*}
- Devyn
- Sure! But what about when \(x=2\)? In this case \[ g(2) = 1\qquad \text {but}\qquad f(2) \text { is undefined!} \]
- Riley
- Right, \(f(2)\) is undefined because we cannot divide by zero. Hmm. Now I see the problem. Yikes!
Suppose \(f\) and \(g\) are functions but the domain of \(f\) is different from the domain
of \(g\). Could it be that \(f\) and \(g\) are actually the same function?
yes no
The domain of a function is part of the “data” of the function. A function is not a
rule for transforming the input to the output, but rather the relationship between a
specified collection of inputs (the domain) and possible outputs (the range).
Let \(f(x) = \sin ^2(x)\) and \(g(u) = \sin ^2(u)\). The domain of each of these functions is all real numbers. Which of the
following statements are true?
2025-01-06 19:54:51 There is not enough information to determine if \(f = g\). The functions are equal. If \(x \neq u\), then \(f \neq g\). We have \(f \neq g\) since \(f\) uses the variable \(x\) and \(g\) uses
the variable \(u\).