Find the domain and range of the following functions: 1. 2/sqrt(x³), (x²+1)/(x³+2), 1/cbrt(x)

Question

ScanSolve AI Tutor · Accepted Answer

We need to find the domain and range for each of the given functions.

1. Function: $f(x) = \frac{2}{\sqrt{x^3}}$

Domain:

For the square root $\sqrt{x^3}$ to be defined, the expression inside must be non-negative, so $x^3 \ge 0$ , which implies $x \ge 0$ .
For the fraction to be defined, the denominator cannot be zero, so $\sqrt{x^3} \ne 0$ , which implies $x^3 \ne 0$ , so $x \ne 0$ .
Combining these conditions, $x$ must be strictly greater than 0.
$$ \text{Domain: } (0, \infty) $$
- Range:
Since $x > 0$ , $x^3$ is always positive, and $\sqrt{x^3}$ is always positive.
Therefore, $f(x) = \frac{2}{\sqrt{x^3}}$ will always be positive.
As $x \to 0^+$ , $\sqrt{x^3} \to 0^+$ , so $f(x) \to \infty$ .
As $x \to \infty$ , $\sqrt{x^3} \to \infty$ , so $f(x) \to 0^+$ . $\text{Range: } (0, \infty)$

2. Function: $g(x) = \frac{x^2+1}{x^3+2}$
- Domain:

For the fraction to be defined, the denominator cannot be zero.

$$
x^3+2 \ne 0
$$
$$
x^3 \ne -2
$$
$$
x \ne \sqrt[3]{-2}
$$
$$
\text{Domain: } (-\infty, -\sqrt[3]{2}) \cup (-\sqrt[3]{2}, \infty)
$$

Range:

As $x \to \infty$ , $g(x) \approx \frac{x^2}{x^3} = \frac{1}{x} \to 0^+$ .
As $x \to -\infty$ , $g(x) \approx \frac{x^2}{x^3} = \frac{1}{x} \to 0^-$ .

Near the vertical asymptote $x = -\sqrt[3]{2}$

As $x \to (-\sqrt[3]{2})^+$, the denominator $x^3+2 \to 0^+$, and the numerator $x^2+1$ is positive, so $g(x) \to \infty$.  
As $x \to (-\sqrt[3]{2})^-$, the denominator $x^3+2 \to 0^-$, and the numerator $x^2+1$ is positive, so $g(x) \to -\infty$.

Since the function approaches both positive and negative infinity, and also approaches 0, it covers all real numbers.
$$ \text{Range: } (-\infty, \infty) $$
3. Function: $h(x) = \frac{1}{\sqrt[3]{x}}$
- Domain:
The cube root $\sqrt[3]{x}$ is defined for all real numbers.
For the fraction to be defined, the denominator cannot be zero, so $\sqrt[3]{x} \ne 0$ , which implies $x \ne 0$ .
$$ \text{Domain: } (-\infty, 0) \cup (0, \infty) $$
- Range:
Let $y = \frac{1}{\sqrt[3]{x}}$ .
We can solve for $x$ in terms of $y$ : $\sqrt[3]{x} = \frac{1}{y}$ $x = \left(\frac{1}{y}\right)^3 = \frac{1}{y^3}$
For $x$ to be a real number, $y$ cannot be zero.
As $x$ can take any non-zero real value, $y$ can also take any non-zero real value.
$$ \text{Range: } (-\infty, 0) \cup (0, \infty) $$
Final Answers:
For $f(x) = \frac{2}{\sqrt{x^3}}$ :
Domain: $\boxed{(0, \infty)}$
Range: $\boxed{(0, \infty)}$

For $g(x) = \frac{x^2+1}{x^3+2}$ :
Domain: $\boxed{(-\infty, -\sqrt[3]{2}) \cup (-\sqrt[3]{2}, \infty)}$
Range: $\boxed{(-\infty, \infty)}$

For $h(x) = \frac{1}{\sqrt[3]{x}}$ :
Domain: $\boxed{(-\infty, 0) \cup (0, \infty)}$
Range: $\boxed{(-\infty, 0) \cup (0, \infty)}$

Find the domain and range of the following functions: 1. 2/sqrt(x³), (x²+1)/(x³+2), 1/cbrt(x)

Step-by-step solution

For the square root $\sqrt{x^3}$ to be defined, the expression inside must be non-negative, so $x^3 \ge 0$ , which implies $x \ge 0$ .

For the fraction to be defined, the denominator cannot be zero, so $\sqrt{x^3} \ne 0$ , which implies $x^3 \ne 0$ , so $x \ne 0$ .

Combining these conditions, $x$ must be strictly greater than 0.

Since $x > 0$ , $x^3$ is always positive, and $\sqrt{x^3}$ is always positive.

Therefore, $f(x) = \frac{2}{\sqrt{x^3}}$ will always be positive.

As $x \to 0^+$ , $\sqrt{x^3} \to 0^+$ , so $f(x) \to \infty$ .

For the fraction to be defined, the denominator cannot be zero.

As $x \to \infty$ , $g(x) \approx \frac{x^2}{x^3} = \frac{1}{x} \to 0^+$ .

As $x \to -\infty$ , $g(x) \approx \frac{x^2}{x^3} = \frac{1}{x} \to 0^-$ .

Near the vertical asymptote $x = -\sqrt[3]{2}$

Since the function approaches both positive and negative infinity, and also approaches 0, it covers all real numbers.

The cube root $\sqrt[3]{x}$ is defined for all real numbers.

For the fraction to be defined, the denominator cannot be zero, so $\sqrt[3]{x} \ne 0$ , which implies $x \ne 0$ .

Let $y = \frac{1}{\sqrt[3]{x}}$ .

For $x$ to be a real number, $y$ cannot be zero.

As $x$ can take any non-zero real value, $y$ can also take any non-zero real value.

Still stuck on this one?

Find the domain and range of the following functions: 1. 2/sqrt(x³), (x²+1)/(x³+2), 1/cbrt(x)

Step-by-step solution

For the square root x3\sqrt{x^3}x3​ to be defined, the expression inside must be non-negative, so x3≥0x^3 \ge 0x3≥0, which implies x≥0x \ge 0x≥0.

For the fraction to be defined, the denominator cannot be zero, so x3≠0\sqrt{x^3} \ne 0x3​=0, which implies x3≠0x^3 \ne 0x3=0, so x≠0x \ne 0x=0.

Combining these conditions, xxx must be strictly greater than 0.

Since x>0x > 0x>0, x3x^3x3 is always positive, and x3\sqrt{x^3}x3​ is always positive.

Therefore, f(x)=2x3f(x) = \frac{2}{\sqrt{x^3}}f(x)=x3​2​ will always be positive.

As x→0+x \to 0^+x→0+, x3→0+\sqrt{x^3} \to 0^+x3​→0+, so f(x)→∞f(x) \to \inftyf(x)→∞.

For the fraction to be defined, the denominator cannot be zero.

As x→∞x \to \inftyx→∞, g(x)≈x2x3=1x→0+g(x) \approx \frac{x^2}{x^3} = \frac{1}{x} \to 0^+g(x)≈x3x2​=x1​→0+.

As x→−∞x \to -\inftyx→−∞, g(x)≈x2x3=1x→0−g(x) \approx \frac{x^2}{x^3} = \frac{1}{x} \to 0^-g(x)≈x3x2​=x1​→0−.

Near the vertical asymptote x=−23x = -\sqrt[3]{2}x=−32​

Since the function approaches both positive and negative infinity, and also approaches 0, it covers all real numbers.

The cube root x3\sqrt[3]{x}3x​ is defined for all real numbers.

For the fraction to be defined, the denominator cannot be zero, so x3≠0\sqrt[3]{x} \ne 03x​=0, which implies x≠0x \ne 0x=0.

Let y=1x3y = \frac{1}{\sqrt[3]{x}}y=3x​1​.

For xxx to be a real number, yyy cannot be zero.

As xxx can take any non-zero real value, yyy can also take any non-zero real value.

Still stuck on this one?

For the square root $\sqrt{x^3}$ to be defined, the expression inside must be non-negative, so $x^3 \ge 0$ , which implies $x \ge 0$ .

For the fraction to be defined, the denominator cannot be zero, so $\sqrt{x^3} \ne 0$ , which implies $x^3 \ne 0$ , so $x \ne 0$ .

Combining these conditions, $x$ must be strictly greater than 0.

Since $x > 0$ , $x^3$ is always positive, and $\sqrt{x^3}$ is always positive.

Therefore, $f(x) = \frac{2}{\sqrt{x^3}}$ will always be positive.

As $x \to 0^+$ , $\sqrt{x^3} \to 0^+$ , so $f(x) \to \infty$ .

As $x \to \infty$ , $g(x) \approx \frac{x^2}{x^3} = \frac{1}{x} \to 0^+$ .

As $x \to -\infty$ , $g(x) \approx \frac{x^2}{x^3} = \frac{1}{x} \to 0^-$ .

Near the vertical asymptote $x = -\sqrt[3]{2}$

The cube root $\sqrt[3]{x}$ is defined for all real numbers.

For the fraction to be defined, the denominator cannot be zero, so $\sqrt[3]{x} \ne 0$ , which implies $x \ne 0$ .

Let $y = \frac{1}{\sqrt[3]{x}}$ .

For $x$ to be a real number, $y$ cannot be zero.

As $x$ can take any non-zero real value, $y$ can also take any non-zero real value.