Even and Odd Function worksheets: Practice and Master Mathematical ... - Free Printable
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Step-by-step solution for: Even and Odd Function worksheets: Practice and Master Mathematical ...
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Step-by-step solution for: Even and Odd Function worksheets: Practice and Master Mathematical ...
It seems like the task involves analyzing or solving problems related to the given functions. However, the problem statement is not explicitly provided in your question. To proceed, I will assume that the task is to analyze the properties of these functions, such as their domains, ranges, symmetry, and behavior. If you have a specific problem or task in mind, please clarify.
1. \( f(x) = x^4 + 5x^2 \)
2. \( f(x) = |x| + x \)
3. \( f(x) = \frac{x}{x^2 + 2} \)
4. \( f(x) = \frac{1}{x - 2} \)
5. \( f(x) = \frac{x^4 + 3}{x^2 + 1} \)
6. \( f(x) = x^2 \sqrt{x^2 + 1} \)
#### 1. \( f(x) = x^4 + 5x^2 \)
- Domain: All real numbers, \( \mathbb{R} \), because it is a polynomial.
- Range: Since \( x^4 \geq 0 \) and \( 5x^2 \geq 0 \) for all \( x \), the function is always non-negative. The minimum value occurs at \( x = 0 \), where \( f(0) = 0 \). As \( x \to \pm\infty \), \( f(x) \to \infty \). Thus, the range is \( [0, \infty) \).
- Symmetry: Even function because \( f(-x) = (-x)^4 + 5(-x)^2 = x^4 + 5x^2 = f(x) \).
- Behavior: The function is increasing for \( x > 0 \) and decreasing for \( x < 0 \).
#### 2. \( f(x) = |x| + x \)
- Domain: All real numbers, \( \mathbb{R} \).
- Range: For \( x \geq 0 \), \( f(x) = x + x = 2x \), so \( f(x) \geq 0 \). For \( x < 0 \), \( f(x) = -x + x = 0 \). Thus, the range is \( [0, \infty) \).
- Symmetry: Not symmetric. It is neither even nor odd.
- Behavior: Constant at 0 for \( x < 0 \) and linearly increasing for \( x \geq 0 \).
#### 3. \( f(x) = \frac{x}{x^2 + 2} \)
- Domain: All real numbers, \( \mathbb{R} \), because the denominator \( x^2 + 2 \neq 0 \) for any \( x \).
- Range: The function is continuous and differentiable. As \( x \to \pm\infty \), \( f(x) \to 0 \). The maximum and minimum values can be found by taking the derivative and setting it to zero. The critical points occur where \( f'(x) = 0 \). After solving, the range is \( \left[ -\frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}} \right] \).
- Symmetry: Odd function because \( f(-x) = \frac{-x}{(-x)^2 + 2} = -\frac{x}{x^2 + 2} = -f(x) \).
- Behavior: The function has a horizontal asymptote at \( y = 0 \).
#### 4. \( f(x) = \frac{1}{x - 2} \)
- Domain: All real numbers except \( x = 2 \), because the denominator cannot be zero. So, the domain is \( \mathbb{R} \setminus \{2\} \).
- Range: All real numbers except 0, because the function can take any value except 0. So, the range is \( \mathbb{R} \setminus \{0\} \).
- Symmetry: Neither even nor odd.
- Behavior: Vertical asymptote at \( x = 2 \) and horizontal asymptote at \( y = 0 \).
#### 5. \( f(x) = \frac{x^4 + 3}{x^2 + 1} \)
- Domain: All real numbers, \( \mathbb{R} \), because the denominator \( x^2 + 1 \neq 0 \) for any \( x \).
- Range: The function is continuous and differentiable. As \( x \to \pm\infty \), \( f(x) \to x^2 \). The minimum value can be found by taking the derivative and setting it to zero. After solving, the minimum value is 3, so the range is \( [3, \infty) \).
- Symmetry: Even function because \( f(-x) = \frac{(-x)^4 + 3}{(-x)^2 + 1} = \frac{x^4 + 3}{x^2 + 1} = f(x) \).
- Behavior: The function is increasing for \( x > 0 \) and decreasing for \( x < 0 \).
#### 6. \( f(x) = x^2 \sqrt{x^2 + 1} \)
- Domain: All real numbers, \( \mathbb{R} \), because the expression under the square root is always non-negative.
- Range: Since \( x^2 \geq 0 \) and \( \sqrt{x^2 + 1} \geq 1 \), the function is always non-negative. The minimum value occurs at \( x = 0 \), where \( f(0) = 0 \). As \( x \to \pm\infty \), \( f(x) \to \infty \). Thus, the range is \( [0, \infty) \).
- Symmetry: Even function because \( f(-x) = (-x)^2 \sqrt{(-x)^2 + 1} = x^2 \sqrt{x^2 + 1} = f(x) \).
- Behavior: The function is increasing for \( x > 0 \) and decreasing for \( x < 0 \).
The analysis of the functions is summarized above. If you have a specific task or problem related to these functions, please provide more details so I can assist further.
\[
\boxed{\text{Analysis provided above}}
\]
Given Functions:
1. \( f(x) = x^4 + 5x^2 \)
2. \( f(x) = |x| + x \)
3. \( f(x) = \frac{x}{x^2 + 2} \)
4. \( f(x) = \frac{1}{x - 2} \)
5. \( f(x) = \frac{x^4 + 3}{x^2 + 1} \)
6. \( f(x) = x^2 \sqrt{x^2 + 1} \)
Analysis of Each Function:
#### 1. \( f(x) = x^4 + 5x^2 \)
- Domain: All real numbers, \( \mathbb{R} \), because it is a polynomial.
- Range: Since \( x^4 \geq 0 \) and \( 5x^2 \geq 0 \) for all \( x \), the function is always non-negative. The minimum value occurs at \( x = 0 \), where \( f(0) = 0 \). As \( x \to \pm\infty \), \( f(x) \to \infty \). Thus, the range is \( [0, \infty) \).
- Symmetry: Even function because \( f(-x) = (-x)^4 + 5(-x)^2 = x^4 + 5x^2 = f(x) \).
- Behavior: The function is increasing for \( x > 0 \) and decreasing for \( x < 0 \).
#### 2. \( f(x) = |x| + x \)
- Domain: All real numbers, \( \mathbb{R} \).
- Range: For \( x \geq 0 \), \( f(x) = x + x = 2x \), so \( f(x) \geq 0 \). For \( x < 0 \), \( f(x) = -x + x = 0 \). Thus, the range is \( [0, \infty) \).
- Symmetry: Not symmetric. It is neither even nor odd.
- Behavior: Constant at 0 for \( x < 0 \) and linearly increasing for \( x \geq 0 \).
#### 3. \( f(x) = \frac{x}{x^2 + 2} \)
- Domain: All real numbers, \( \mathbb{R} \), because the denominator \( x^2 + 2 \neq 0 \) for any \( x \).
- Range: The function is continuous and differentiable. As \( x \to \pm\infty \), \( f(x) \to 0 \). The maximum and minimum values can be found by taking the derivative and setting it to zero. The critical points occur where \( f'(x) = 0 \). After solving, the range is \( \left[ -\frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}} \right] \).
- Symmetry: Odd function because \( f(-x) = \frac{-x}{(-x)^2 + 2} = -\frac{x}{x^2 + 2} = -f(x) \).
- Behavior: The function has a horizontal asymptote at \( y = 0 \).
#### 4. \( f(x) = \frac{1}{x - 2} \)
- Domain: All real numbers except \( x = 2 \), because the denominator cannot be zero. So, the domain is \( \mathbb{R} \setminus \{2\} \).
- Range: All real numbers except 0, because the function can take any value except 0. So, the range is \( \mathbb{R} \setminus \{0\} \).
- Symmetry: Neither even nor odd.
- Behavior: Vertical asymptote at \( x = 2 \) and horizontal asymptote at \( y = 0 \).
#### 5. \( f(x) = \frac{x^4 + 3}{x^2 + 1} \)
- Domain: All real numbers, \( \mathbb{R} \), because the denominator \( x^2 + 1 \neq 0 \) for any \( x \).
- Range: The function is continuous and differentiable. As \( x \to \pm\infty \), \( f(x) \to x^2 \). The minimum value can be found by taking the derivative and setting it to zero. After solving, the minimum value is 3, so the range is \( [3, \infty) \).
- Symmetry: Even function because \( f(-x) = \frac{(-x)^4 + 3}{(-x)^2 + 1} = \frac{x^4 + 3}{x^2 + 1} = f(x) \).
- Behavior: The function is increasing for \( x > 0 \) and decreasing for \( x < 0 \).
#### 6. \( f(x) = x^2 \sqrt{x^2 + 1} \)
- Domain: All real numbers, \( \mathbb{R} \), because the expression under the square root is always non-negative.
- Range: Since \( x^2 \geq 0 \) and \( \sqrt{x^2 + 1} \geq 1 \), the function is always non-negative. The minimum value occurs at \( x = 0 \), where \( f(0) = 0 \). As \( x \to \pm\infty \), \( f(x) \to \infty \). Thus, the range is \( [0, \infty) \).
- Symmetry: Even function because \( f(-x) = (-x)^2 \sqrt{(-x)^2 + 1} = x^2 \sqrt{x^2 + 1} = f(x) \).
- Behavior: The function is increasing for \( x > 0 \) and decreasing for \( x < 0 \).
Final Answer:
The analysis of the functions is summarized above. If you have a specific task or problem related to these functions, please provide more details so I can assist further.
\[
\boxed{\text{Analysis provided above}}
\]
Parent Tip: Review the logic above to help your child master the concept of even and odd functions worksheet pdf.