Exponential Functions

Approach:

For an exponential function $y = a^x$ with $a > 0$ the base is positive, so every real exponent is meaningful; because a positive base raised to any real power stays positive, the function never takes the value $0$ or a negative value.

Steps:

1. For every $x \in \mathbb{R}$, since $a > 0$ the real power $a^x$ is well defined and positive, so the domain is $\mathbb{R}$.
2. The inequality $a^x > 0$ always holds, so all function values are positive and the range contains neither $0$ nor negative numbers.
3. As $x \to +\infty$ we have $a^x \to +\infty$, while as $x \to -\infty$ we have $a^x \to 0^+$ without ever reaching $0$, so the range is $(0, +\infty)$.

Answer:

Domain: $\mathbb{R}$; Range: $(0, +\infty)$.

Approach:

The criterion for an exponential function is that it can be written as $y = a^x$ with $a > 0$ and $a \neq 1$, that is, the variable appears only in the exponent; match each of the four functions against that form.

Steps:

1. $y = 3^x$ is an exponential function with base $a = 3$, which satisfies $a > 0$ and $a \neq 1$.
2. In $y = x^3$ the variable sits in the base position, so this is a power function rather than an exponential function.
3. $y = 2^{-x} = (1/2)^x$ has base $1/2 > 0$ with $1/2 \neq 1$, so it is still an exponential function.
4. $y = e^x$ is an exponential function with the natural constant $e$ as base.

Answer:

$y = 3^x$, $y = 2^{-x}$, $y = e^x$ are exponential functions, $y = x^3$ is not.

Approach:

Use the two key properties of the exponential function: the graph passes through the fixed point $(0, 1)$, and it increases when the base is above $1$ and decreases when the base lies between $0$ and $1$; that fixes the shape of both curves and allows the comparison.

Steps:

1. Both curves pass through $(0, 1)$, the common feature of all exponential functions.
2. For $y = 2^x$ the base $2 > 1$, so the function is monotonically increasing on $\mathbb{R}$; the graph approaches the $x$-axis on the left and grows without bound on the right.
3. For $y = (1/2)^x$ the base $0 < 1/2 < 1$, so the function is monotonically decreasing on $\mathbb{R}$; the graph grows without bound on the left and approaches the $x$-axis on the right.
4. Since $(1/2)^x = 2^{-x}$, the two graphs are symmetric about the $y$-axis and their directions of monotonicity are opposite.

Answer:

$y = 2^x$ is monotonically increasing, $y = (1/2)^x$ is monotonically decreasing.

Approach:

An asymptote is a line the graph approaches arbitrarily closely without ever meeting it; for $y = a^x$ with $a > 1$ it is enough to study the behaviour of the function as $x$ tends to negative infinity.

Steps:

1. As $x \to -\infty$, $a^x = \frac{1}{a^{-x}} \to 0^+$.
2. As $x$ moves arbitrarily far to the left the graph hugs the $x$-axis, while the function value stays positive and never crosses that axis.
3. Therefore the $x$-axis, that is $y = 0$, is the horizontal asymptote of $y = a^x$ for $a > 1$.

Answer:

The x-axis ($y=0$) is its asymptote.

Approach:

To decide whether a function is exponential, check only whether it can be written as $y = a^x$ with a base satisfying $a > 0$ and $a \neq 1$.

Steps:

1. The base of $y = (-2)^x$ is $-2$, which does not satisfy the requirement $a > 0$.
2. For instance at $x = \frac{1}{2}$ the value $(-2)^{1/2} = \sqrt{-2}$ has no meaning over the reals; in general an arbitrary real power of a negative base cannot be given a real meaning on all of $\mathbb{R}$.
3. Since the base of an exponential function must be positive, $(-2)^x$ cannot be treated as an exponential function.

Answer:

No. The base $a$ of an exponential function must be greater than 0 and not equal to 1.

Approach:

Recover the base from the given function values: for an exponential function $y = a^x$ the value at $x = 0$ is always $1$, and the value at $x = 1$ is exactly the base $a$ itself.

Steps:

1. At $x = 0$ we have $y(0) = a^0 = 1$, which agrees with the given $y(0) = 1$ and confirms that the base satisfies $a > 0$ and $a \neq 1$.
2. At $x = 1$ we have $y(1) = a^1 = a$.
3. The condition $y(1) = 3$ gives $a = 3$, and $3 > 0$ with $3 \neq 1$ meets the requirement on the base of an exponential function.

Answer:

$y(1) = a^1 = 3$, so $a = 3$.

Approach:

Check for each option whether it can be written as $y = a^x$ with $a > 0$ and $a \neq 1$, that is, whether the variable occurs only in the exponent.

Steps:

1. (A) $y = 2^x$ has base $2 > 0$ with $2 \neq 1$, so it is an exponential function.
2. (B) $y = x^2$ has the variable in the base position, so it is a power function and not an exponential function.
3. (C) $y = e^x$ has base the natural constant $e > 0$ with $e \neq 1$, so it is an exponential function.
4. (D) $y = 2^{-x} = (1/2)^x$ has base $1/2 > 0$ with $1/2 \neq 1$, so it is an exponential function.

Answer:

(A), (C), (D) are exponential functions, (B) is not.

Approach:

The monotonicity of an exponential function is decided solely by how the base compares with $1$: a base above $1$ gives an increasing function, a base between $0$ and $1$ gives a decreasing one.

Steps:

1. For any $x_1 < x_2$, since $0 < a < 1$ the powers decrease as the exponent grows, that is $a^{x_1} > a^{x_2}$.
2. Hence $y = a^x$ is monotonically decreasing on $\mathbb{R}$.
3. Accordingly the graph grows without bound on the left and approaches the $x$-axis on the right, mirroring the case $a > 1$ about the $y$-axis.

Answer:

$y$ is monotonically decreasing.