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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?

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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?

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I took a closer look at the method for determining the range of a composite function described on the U of T website and I came up with the same answer as the one already posted (unfortunately I can't fully follow the reasoning in that solution, even though it seems to be correct).

The method described on the U of T website states that to find the range of a composite function, we must first find the range of the outer function. Then we determine the range of the inner function, which we can also regard as the domain of the outer function in the composition, and we check whether this domain eliminates any of the values from the range of the outer function. Below is the application of this method to the given problem.


Given $f(x) = \dfrac{1}{x - 3}$ and $g(x) = \sqrt{x}$, we want to find the range of:

$$(f \circ g)(x)$$

The range of the outer function, $f(x)$, is:

$$ \{ y \in \mathbb{R} \mid y \neq 0\}$$

The range of the inner function, $g(x)$, is:

$$ \{ y \in \mathbb{R} \mid y \geq 0\}$$

Because the range of the inner function becomes the domain of the outer function when we consider the composite function $f(g(x)) = \dfrac{1}{\sqrt{x} - 3}$, we must eliminate values such as $-\dfrac{1}{4}, -\dfrac{1}{5}, -\dfrac{1}{6}$, etc. More precisely, the range of the composite function can't be in the interval $\left(-\dfrac{1}{3}, 0\right)$. We also determined that $0$ is not in the range of the outer function, so actually the exclusionary interval should be $\left(-\dfrac{1}{3}, 0\right]$. The range of the composite function is the complement of this interval, which is:

$$\biggl( - \infty, -\dfrac{1}{3} \biggr] \cup \biggl( 0, \infty \biggr) $$

or, equivalently:

$$ \left\{ y \in \mathbb{R} \mid y \leq - \dfrac{1}{3} \ \ \ \wedge \ \ \ y > 0 \right\}$$


(One of my difficulties with this question was realizing how the range of the inner function affected that of the outer. Drawing a number line helped a lot. I was also confused by the fractional values, until I started listing specific ones, and realizing they are all smaller than $-1 / 3$ and that they all fell within a specific interval).

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