https://file.aiquickdraw.com/mathbot/file/1729484174839efgwtpq5.jpg

The function given is:

\[
f(x) = \frac{x^3}{x^2 + 1}
\]

To sketch the graph of this function, we will go through the following steps:

### 1. **Domain**:
Since \(x^2 + 1 > 0\) for all real \(x\), the function is defined for all real values of \(x\). So, the domain of \(f(x)\) is \( (-\infty, \infty) \).

### 2. **Symmetry**:
Let's check for symmetry:
- **Even function**: A function is even if \(f(-x) = f(x)\). Here, \(f(-x) = \frac{(-x)^3}{(-x)^2 + 1} = \frac{-x^3}{x^2 + 1} \neq f(x)\), so the function is not even.
- **Odd function**: A function is odd if \(f(-x) = -f(x)\). Since \(f(-x) = -f(x)\), the function is odd. Therefore, the graph is symmetric about the origin.

### 3. **Intercepts**:
- **x-intercept**: Set \(f(x) = 0\), which implies that \(x^3 = 0\), so \(x = 0\). Therefore, the x-intercept is at \((0, 0)\).
- **y-intercept**: At \(x = 0\), \(f(0) = 0\), so the y-intercept is also at \((0, 0)\).

### 4. **Asymptotes**:
- **Horizontal asymptote**: As \(x \to \infty\), \(f(x) \approx \frac{x^3}{x^2} = x\), and as \(x \to -\infty\), \(f(x) \approx \frac{x^3}{x^2} = x\). Therefore, there is no horizontal asymptote, but as \(x\) becomes very large, the function grows linearly with \(x\), indicating an oblique asymptote.
  
### 5. **Critical Points and Monotonicity**:
To find critical points, we take the derivative of \(f(x)\):
\[
f'(x) = \frac{(3x^2)(x^2 + 1) - x^3(2x)}{(x^2 + 1)^2} = \frac{3x^2(x^2 + 1) - 2x^4}{(x^2 + 1)^2} = \frac{x^2(3 - x^2)}{(x^2 + 1)^2}
\]
The critical points occur where \(f'(x) = 0\), which happens when \(x = -\sqrt{3}, 0, \sqrt{3}\). 

- For \(x = 0\), we already know that \(f(0) = 0\).
- For \(x = \sqrt{3}\), we can calculate \(f(\sqrt{3})\) and similarly for \(x = -\sqrt{3}\).

The sign of \(f'(x)\) tells us that the function increases on \((- \infty, -\sqrt{3})\), decreases on \((- \sqrt{3}, 0)\), increases on \((0, \sqrt{3})\), and then decreases again on \((\sqrt{3}, \infty)\).

### 6. **Concavity and Inflection Points**:
To determine concavity, we calculate the second derivative \(f''(x)\) and look for changes in sign.

### 7. **Sketching the Graph**:
- The function is odd, so it is symmetric about the origin.
- There are no vertical asymptotes, and the function has an oblique asymptote at large values of \(x\).
- It has critical points at \(x = -\sqrt{3}\), \(x = 0\), and \(x = \sqrt{3}\).

Would you like to proceed with plotting this graph or need further details on any step?

Here are 5 related questions to deepen your understanding:

1. How do you compute the second derivative to determine concavity?
2. What is the significance of critical points in a function's graph?
3. How does the odd symmetry affect the shape of the graph?
4. Can we estimate the value of \(f(x)\) for large values of \(x\)?
5. How do asymptotes guide the behavior of the function at extremes?

**Tip**: Always consider the symmetry of the function early when analyzing or sketching a graph—it can save you time in predicting its shape!

Sketch the graph of f(x) = x^3 / (x^2 + 1)

Learn how to sketch the graph of f(x) = x^3 / (x^2 + 1). This solution includes domain analysis, symmetry, intercepts, asymptotes, critical points, and concavity. A detailed calculus-based approach is used, suitable for advanced high school students or first-year college calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation