Domain and Range of Exponential Functions

Q: Why is the range $y > 0$ instead of $y \geq 0$?

Because the curve never actually reaches a height of $0$. Every output of $y = 2^x$ is positive, even for very negative inputs: $2^{-10} = \dfrac{1}{1024}$, which is tiny but still above zero. A boundary value belongs in the range only when the graph truly hits it.

Q: Is the domain ever something other than all real numbers?

Not for an exponential of the form $y = a \cdot b^x$. Every real exponent produces an output, so nothing is excluded. A restriction would have to come from context, such as a population model where only whole numbers of years make sense.

Domain asks which inputs a function will accept. Range asks which outputs it actually produces. For exponential functions the domain answer is the same every time, and the range answer always comes from one feature of the graph.

That feature is the horizontal asymptote, the flat line the curve gets closer and closer to without ever landing on it. Because the curve approaches that height but never reaches it, the range is written with a strict inequality rather than $\geq$ or $\leq$ .

The domain is always all real numbers

In $y = a \cdot b^x$ , the input $x$ sits in the exponent. There is no exponent an exponential function refuses. Negative inputs are fine, since $3^{-2} = \dfrac{1}{9}$ . Zero is fine, since $3^0 = 1$ . Fractions are fine, since $\displaystyle 3^{\frac{1}{2}} = \sqrt{3}$ .

Nothing gets divided by zero and nothing goes under a square root, which are the two usual reasons a domain gets restricted. So the domain of every exponential function of this form is all real numbers, written $(-\infty, \infty)$ in interval notation.

The asymptote decides the range

Look at $y = 3^x$ . As $x$ moves left, the outputs shrink: $3^{-2} = \dfrac{1}{9}$ , then $3^{-5} = \dfrac{1}{243}$ , then smaller still. They keep halving toward the $x$ -axis, but no exponent ever makes the output exactly $0$ . The line $y = 0$ is the horizontal asymptote.

Since $0$ is approached and never reached, $0$ is not an output, so it stays out of the range. That is why you write $y > 0$ and not $y \geq 0$ . Compare it to a parabola with vertex $(0, 2)$ , which really does touch its lowest height and therefore gets $y \geq 2$ . Reaching means include, approaching means exclude.

Which side of the asymptote, and interval notation

The whole curve stays on one side of the asymptote. If the graph sits above $y = 0$ , every output is positive and the range is $y > 0$ . If the graph has been reflected so it sits below $y = 0$ , every output is negative and the range is $y < 0$ . Growth or decay does not matter here. A decaying curve like $y = 5(0.4)^x$ still lives entirely above the $x$ -axis.

Interval notation says the same thing with brackets. A round bracket leaves the endpoint out, and a square bracket includes it. Since the asymptote height is never an output, it always gets a round bracket: $y > 0$ becomes $(0, \infty)$ and $y < 0$ becomes $(-\infty, 0)$ . Infinity always takes a round bracket too, because it is not a number the graph reaches.

Worked examples

Example 1: a growing exponential

Find the domain and range of $y = 3^x$ .

Any exponent is allowed, so the domain is unrestricted

\text{domain: all real numbers}

Test a few inputs to see the outputs stay positive

3^{-2} = \dfrac{1}{9}, \quad 3^0 = 1, \quad 3^2 = 9

The curve approaches

y = 0

but never reaches it

\text{asymptote: } y = 0

Answer: Domain: all real numbers. Range: $y > 0$ .

Example 2: decay does not change the range

Find the range of $y = 5(0.4)^x$ .

Check outputs on both sides of zero

5(0.4)^{-1} = 12.5, \quad 5(0.4)^0 = 5, \quad 5(0.4)^2 = 0.8

The outputs shrink toward the

x

-axis but stay positive

\text{asymptote: } y = 0

The curve is entirely above the asymptote

y > 0

Answer: Range: $y > 0$ . The curve falls instead of rising, but it never crosses the $x$ -axis.

Example 3: a curve below the asymptote

Find the range of $y = -2(0.5)^x$ .

Check a few outputs

-2(0.5)^{-2} = -8, \quad -2(0.5)^0 = -2, \quad -2(0.5)^2 = -0.5

Every output is negative, so the curve sits below the

x

-axis

\text{asymptote: } y = 0

The outputs climb toward

0

without arriving

y < 0

Answer: Range: $y < 0$ .

Example 4: writing the answers in interval notation

Write the domain $x =$ all real numbers and the range $y > 0$ in interval notation.

The domain has no endpoints at all

(-\infty, \infty)

The range starts at

0

, which is excluded, so use a round bracket

y > 0

There is no upper bound, and infinity always gets a round bracket

(0, \infty)

Answer: Domain: $(-\infty, \infty)$ . Range: $(0, \infty)$ .

Try one yourself

Common questions

Why is the range

y > 0

instead of

y \geq 0

Because the curve never actually reaches a height of $0$ . Every output of $y = 2^x$ is positive, even for very negative inputs: $2^{-10} = \dfrac{1}{1024}$ , which is tiny but still above zero. A boundary value belongs in the range only when the graph truly hits it.

Is the domain ever something other than all real numbers?

Not for an exponential of the form $y = a \cdot b^x$ . Every real exponent produces an output, so nothing is excluded. A restriction would have to come from context, such as a population model where only whole numbers of years make sense.

How do I know whether the range is

y > 0

y < 0

Look at which side of the asymptote the graph sits on, or evaluate the function once. If the output is positive, the whole curve is above $y = 0$ and the range is $y > 0$ . If the output is negative, the whole curve is below and the range is $y < 0$ . The curve never crosses the asymptote, so one test point settles it.

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