Range

Take $f(x) = x^2$. Try to find an input that produces the output $-1$. Write down what happens when you try to solve $x^2 = -1$, and say what that tells you about which numbers are in the range.

The range of a function is the set of output values it actually produces as the input varies over the whole domain. On a graph, the domain is read across and the range is read up.

On a graph, each point is an input paired with its output. The input is the horizontal coordinate and the output is the vertical one. So:

the domain is the set of horizontal positions the graph occupies, found by looking along the horizontal axis;
the range is the set of heights the graph occupies, found by looking up and down the vertical axis.

A useful way to see it: shine a light straight down onto the horizontal axis and the shadow of the graph is the domain. Shine it sideways onto the vertical axis and the shadow is the range.

For $y = x^2 - 4$ the curve extends left and right forever, so the domain is all real numbers. It has a lowest point at $(0,-4)$ and rises without limit on both sides, so its heights are everything from $-4$ upward:

$$\text{range} = \{y : y \geq -4\} = [-4, \infty)$$

The value $-4$ is included because the curve actually reaches it, at $x = 0$. That check, whether the boundary is attained, decides between a square bracket and a round one.

A domain can be found by inspection: look for a denominator, look for a root, and you are done. A range cannot.

Compare $f(x) = x^2 - 4$ with $g(x) = x^3 - 4$. Both are polynomials, so both have domain all real numbers, and the two rules look almost identical. But:

$$f \text{ has range } [-4, \infty) \qquad g \text{ has range } (-\infty, \infty)$$

The cube can produce any real number, including large negatives, because an odd power keeps the sign of its input. The square cannot go below zero, so $x^2 - 4$ cannot go below $-4$.

Nothing on the surface of the two rules distinguishes them. You have to know what the function does, which in practice means sketching it or reasoning about its shape.

Linear, $f(x) = mx + c$ with $m \neq 0$. The line rises or falls without limit in both directions, so the range is all real numbers. If $m = 0$ the function is constant and its range is the single value $\{c\}$.

Quadratic, $f(x) = ax^2 + bx + c$. The vertex is the boundary. If $a > 0$ the parabola opens upward and the vertex is a minimum, so the range is $[k, \infty)$ where $k$ is the vertex height. If $a < 0$ it opens downward and the range is $(-\infty, k]$.

Cubic, and any odd-degree polynomial. The two ends go opposite ways without limit, and the graph is unbroken, so every real value is attained: the range is all real numbers.

Square root, $f(x) = \sqrt{x}$. The symbol means the non-negative root, so outputs are never negative, and every non-negative value is reached. The range is $[0, \infty)$.

Reciprocal, $f(x) = \dfrac{1}{x}$. No input produces the output $0$, since $\dfrac{1}{x} = 0$ has no solution. Every other real value is produced. The range is all real numbers except $0$.

That last one is worth pausing on. The horizontal axis is an asymptote: the curve gets arbitrarily close to height zero but never lands on it.

The range is defined in terms of the domain, so changing one changes the other.

Take $f(x) = x^2$ on three different domains.

On all real numbers, the range is $[0,\infty)$: every non-negative value is reached.

On $x \geq 0$, the range is still $[0,\infty)$. Cutting away the negative inputs removed nothing from the outputs, because each of those outputs was also produced by a positive input.

On $1 \leq x \leq 3$, the range is $[1,9]$: the smallest output is $1^2 = 1$ and the largest is $3^2 = 9$, and everything between is reached.

The middle case is the instructive one. Shrinking the domain does not always shrink the range, because a many-to-one function may reach the same output from more than one input. Removing one of them leaves the output still available.

When a domain is restricted to an interval, check the two endpoints and any turning point inside the interval. On $-1 \leq x \leq 3$ the endpoints give $1$ and $9$, but the vertex at $x = 0$ gives $0$, which is lower than both. The range is $[0,9]$, and the endpoints alone would have missed it.

Five items on ranges. Work each one, then reveal the answer.

1. 1 State the range of $f(x) = 3x + 1$.

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   A non-horizontal line reaches every height.All real numbers

2. 2 State the range of $f(x) = x^2 + 7$.

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   The square is at least zero, so add $7$.$y \geq 7$

3. 3 State the range of $f(x) = -x^2$.

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   The square is at least zero, so its negative is at most zero.$y \leq 0$

4. 4 State the range of $f(x) = \sqrt{x} - 5$.

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   The root is at least zero, so subtract $5$.$y \geq -5$

5. 5 State the range of $f(x) = \dfrac{1}{x}$.

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   No input gives the output zero.All real numbers except $0$