Power Functions

Approach:

The domain of a power function $y = x^a$ depends on the exponent $a$; for a positive integer exponent both the domain and the range are $\mathbb{R}$, and the parity is decided by the parity of the exponent.

Steps:

1. Since $a = 3$ is a positive integer, $x^3$ is meaningful for every real $x$, so the domain is $\mathbb{R}$.
2. The cubic function is monotonically increasing and unbounded on $\mathbb{R}$ and its graph passes through the origin, so the range is $\mathbb{R}$.
3. Computing $f(-x) = (-x)^3 = -x^3 = -f(x)$, that is $f(-x) = -f(x)$, so $y = x^3$ is an odd function.

Answer:

Domain: $\mathbb{R}$; Range: $\mathbb{R}$; $y = x^3$ is an odd function.

Approach:

The standard shape of a power function is $y = x^a$ with $a$ constant, where the base is the variable $x$ itself and the coefficient is a constant; compare each of the four expressions with that shape.

Steps:

1. $y = 2x^2$ has coefficient $2$ instead of $1$, so it cannot be written in the form $x^a$ and is not a power function.
2. $y = 3^x$ has the constant base $3$ with the variable in the exponent position, so it is an exponential function and not a power function.
3. $y = x^{-1}$ matches $y = x^a$ with $a = -1$, so it is a power function.
4. $y = \sqrt{x}$ can be written as $y = x^{1/2}$, so it is a power function as well.

Answer:

The power functions are $y = x^{-1}$ and $y = \sqrt{x}$ (i.e., $y = x^{1/2}$). Neither $y = 2x^2$ nor $y = 3^x$ is a power function.

Approach:

The quadratic and the cubic have different graph shapes: the first is an upward-opening parabola symmetric about the $y$-axis, the second a curve symmetric about the origin; use these symmetries to compare the behaviour on the two half-lines.

Steps:

1. The graph of $y = x^2$ is an upward-opening parabola through $(0, 0)$ and symmetric about the $y$-axis, so $y = x^2$ is an even function.
2. On $x > 0$ the value of $x^2$ grows as $x$ grows, so $y = x^2$ is monotonically increasing there; on $x < 0$ the value of $|x|$ shrinks as $x$ grows towards $0$, so $y = x^2$ is monotonically decreasing there.
3. The graph of $y = x^3$ passes through $(0, 0)$ and is symmetric about the origin, and it is monotonically increasing over the whole real line, so $y = x^3$ is an odd function.
4. Both curves pass through the origin, but the parabola turns back on the negative half-line while the cubic keeps rising, which is exactly the difference between their parities.

Answer:

$y = x^2$ is monotonically increasing when $x>0$ and monotonically decreasing when $x<0$, and it is an even function; $y = x^3$ is monotonically increasing over the entire real numbers, and it is an odd function.

Approach:

A negative exponent means a reciprocal: $x^{-2} = \frac{1}{x^2}$, and that representation settles both the domain and the range at once.

Steps:

1. From $x^{-2} = \frac{1}{x^2}$ the denominator must not vanish, so $x \neq 0$ and the domain is $\mathbb{R} \setminus \{0\}$.
2. The square of any non-zero real number is positive, so $\frac{1}{x^2} > 0$ and all function values are positive.
3. As $x \to 0$ we have $\frac{1}{x^2} \to +\infty$, while as $x \to \pm\infty$ we have $\frac{1}{x^2} \to 0^+$ without ever reaching $0$, so the range is $(0, +\infty)$.

Answer:

Domain: $x \neq 0$; Range: $(0, +\infty)$.

Approach:

To decide whether a function is a power function, check only whether it can be written as $y = x^a$ with $a$ constant; the piecewise structure of the absolute value does not fit that form.

Steps:

1. The absolute value function equals $x$ for $x \geq 0$ and $-x$ for $x < 0$, so it is defined piecewise.
2. If a constant $a$ made $x^a = |x|$ hold for every $x$, then on $x > 0$ we would get $a = 1$, but then $x^a = x$ takes negative values on $x < 0$, contradicting $|x| > 0$.
3. Hence $|x|$ cannot be written in the form $x^a$ with $a$ constant.

Answer:

No, it is not a power function. $|x|$ cannot be expressed in the form $x^a$, where $a$ is a constant.

Approach:

Symmetry of the graph about the origin is equivalent to the function being odd, so compare $f(-x)$ with $f(x)$.

Steps:

1. Compute $f(-x) = (-x)^{1/3} = \sqrt[3]{-x} = -\sqrt[3]{x} = -f(x)$.
2. Since $f(-x) = -f(x)$, the function $y = x^{1/3}$ is an odd function.
3. The graph of an odd function is symmetric about the origin: rotating it by $180$ degrees about the origin maps it onto itself.

Answer:

$y = x^{1/3}$ is an odd function, and its graph is symmetric about the origin.

Approach:

The monotonicity of a power function on the positive half-line is fixed by the sign of the exponent $a$: a positive exponent gives an increasing function, a negative exponent a decreasing one.

Steps:

1. When $a > 0$, for any $0 < x_1 < x_2$ we have $x_1^a < x_2^a$.
2. Therefore $y = x^a$ is monotonically increasing on $(0, +\infty)$.
3. The graph passes through $(1, 1)$, tends to $0$ as $x \to 0^+$ and tends to $+\infty$ as $x \to +\infty$.

Answer:

When $x>0$, $y = x^a$ is monotonically increasing.

Approach:

The reciprocal function $y = x^{-1} = \frac{1}{x}$ has its domain split by $x = 0$ into two intervals, so the monotonicity has to be discussed separately on each of them.

Steps:

1. From $x^{-1} = \frac{1}{x}$ the domain is $x \neq 0$, that is the two intervals $(-\infty, 0)$ and $(0, +\infty)$.
2. On $(0, +\infty)$ the denominator grows while the numerator stays $1$, so $\frac{1}{x}$ decreases and the function is monotonically decreasing.
3. On $(-\infty, 0)$ we have $\frac{1}{x} = -\frac{1}{|x|}$, so the value again decreases as $|x|$ grows, and the function is monotonically decreasing there as well.

Answer:

When $x>0$, $y = x^{-1}$ is monotonically decreasing; when $x<0$, $y = x^{-1}$ is monotonically increasing.

Approach:

The value of the zeroth power has to be separated from the undefined case $0^0$: as long as $x \neq 0$ we have $x^0 = 1$, and then compare with the form $y = x^a$.

Steps:

1. For any $x \neq 0$ the identity $x^a \cdot x^{-a} = x^0 = 1$ shows that the zeroth power is constantly $1$; at $x = 0$ the expression $0^0$ is undefined, so the domain is $x \neq 0$.
2. The function value is always $1$, so $y = x^0$ is a constant function.
3. It can be written as $y = x^a$ by taking $a = 0$, which matches the definition of a power function, so it also belongs to the power functions.

Answer:

$y = x^0 = 1$ (where $x \neq 0$), is a constant function, and also belongs to power functions.

Approach:

Compare $f(-x)$ with $f(x)$; the sign change is decided by the parity of the exponent $a$.

Steps:

1. For an integer exponent $a$ we have $f(-x) = (-x)^a = (-1)^a x^a$.
2. When $a$ is even, $(-1)^a = 1$, so $f(-x) = f(x)$ and $y = x^a$ is an even function.
3. When $a$ is odd, $(-1)^a = -1$, so $f(-x) = -f(x)$ and $y = x^a$ is an odd function.

Answer:

When $a$ is even, $y = x^a$ is an even function; when $a$ is odd, $y = x^a$ is an odd function.

Approach:

Use one of the given function values to write down an equation for the exponent $p$ and solve it, then check the result against the other function value.

Steps:

1. From $f(2) = 2^p = 8 = 2^3$, comparing equal bases gives $p = 3$.
2. Checking the other condition, $f(4) = 4^3 = 64$, which agrees with the given $f(4) = 64$.
3. The two conditions confirm each other, so $p = 3$ is indeed the required exponent.

Answer:

From $f(2) = 2^p = 8$, we get $p = 3$. Verification: $f(4) = 4^3 = 64$, which holds.

Approach:

Check each option against the form $y = x^a$ with $a$ constant, which separates power functions from exponential and logarithmic ones.

Steps:

1. (A) $y = x^{1/2}$ matches the form $y = x^a$, so it is a power function.
2. (B) $y = 2^x$ has the constant base $2$ with the variable in the exponent position, so it is an exponential function.
3. (C) $y = x^{-3}$ matches the form $y = x^a$, so it is a power function.
4. (D) $y = \ln x$ has its argument inside a logarithm, so it is a logarithmic function.

Answer:

(A) and (C) belong to power functions, (B) is an exponential function, (D) is a logarithmic function.

Approach:

Rewrite $x^a$ in reciprocal form $x^a = \frac{1}{x^{-a}}$ and use the fact that taking a reciprocal reverses the order of two positive numbers.

Steps:

1. When $a < 0$ we have $-a > 0$, so $x^{-a}$ is monotonically increasing on $(0, +\infty)$.
2. Taking the reciprocal reverses the inequality: if $x_1 < x_2$ then $x_1^{-a} < x_2^{-a}$, hence $\frac{1}{x_1^{-a}} > \frac{1}{x_2^{-a}}$.
3. Therefore $y = x^a$ is monotonically decreasing on $(0, +\infty)$, and its graph passes through $(1, 1)$.

Answer:

It is monotonically decreasing on $(0, +\infty)$.

Approach:

The value of the zeroth power has to be separated from the undefined case $0^0$: as long as $x \neq 0$ we have $x^0 = 1$, and then compare with the form $y = x^a$ to give the reason.

Steps:

1. For any $x \neq 0$ the zeroth power is constantly $1$; at $x = 0$ the expression $0^0$ is undefined, so the domain is $x \neq 0$.
2. The function value is always $1$, so $y = x^0$ is a constant function.
3. It can be written as $y = x^a$ by taking $a = 0$, which matches the definition of a power function, so it belongs to the power functions.

Answer:

$y = x^0 = 1$ (where $x \neq 0$), is a constant function, and also belongs to power functions.

Approach:

The symmetry of the graph is decided by the parity of the function, so first use $f(-x) = (-1)^a f(x)$ to classify the function and then state the symmetry.

Steps:

1. For a non-zero integer $a$ we have $f(-x) = (-x)^a = (-1)^a x^a$.
2. When $a$ is odd, $(-1)^a = -1$, so $f(-x) = -f(x)$ and $y = x^a$ is an odd function.
3. The graph of an odd function is symmetric about the origin: rotating it by $180$ degrees about the origin maps it onto itself.

Answer:

It is an odd function, and its graph is symmetric about the origin.