I am attempting a Functions and Inverses self-test offered by the University of Toronto, and I'm trying to understand why my answer for question (4) differs from the given one.
Given $f(x) = \dfrac{1}{x – 3}$ and $g(x) = \sqrt{x}$, we are asked to find the domain and range of the combined function
$$(f \circ g)(x)$$
My solution for the domain matches the given one, and I won't bother reproducing it here, but my solution for the range does not. This is how I determined the range:
Since $(f \circ g)(x) = f(g(x)) = \dfrac{1}{\sqrt{x} – 3}$, it's easy to see that $y \neq 0$, since the numerator isn't $0$. We also know that $\sqrt{x} \geq 0$, which in turn implies that $y \geq – \dfrac{1}{3}$.
Combining these two restrictions, my solution for the range is
$$\{y \in \mathbb{R} \mid y \geq – \dfrac {1}{3} \wedge y \neq 0 \}$$
The given solution, however, is:
$$\{y \in \mathbb{R} \mid y \neq > 0 \}$$
I'm not sure what the $\neq >$ notation means. I'm assuming it's a typo, and it's actually supposed to be just a $\neq$ sign. But even so, why isn't the $y \geq -\dfrac{1}{3}$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?
Best Answer
The range is $(0,\infty) \cup (-\infty, -\frac 1 3]$. To see this write the range as $\{\frac 1 {t-3}: t \geq 0, t \neq 3\}$. Find $\{\frac 1 {t-3}:0 \leq t < 3\}$ and $\{\frac 1 {t-3}: 3 < t <\infty)\}$ separately. These can be written as $\{\frac 1 s:-3 \leq s < 0\}$ and $\{\frac 1 s: 0 < s <\infty)\}$. Can you compute the range now?