Math Problem Statement
Solution
(b) Domain and Range of the Function
- Domain: The function is defined for all values of except where the denominator is zero. Thus, the domain is (all real numbers except ).
- Range: Since is always positive (being the reciprocal of a squared term), the range is . Therefore, the range is .
(c) Re-write the Function with Argument
If we replace with , the function becomes:
(d) Use the Graph to Estimate
From the graph, we can observe that at , the value of appears to be around .
(e) Calculate Exactly
To find exactly:
(f) Modify the Function to Exclude from the Domain
To omit from the domain, we can define the function as:
\frac{1}{(x - 3)^2}, & x \neq 3 \text{ and } x \neq 4 \\ \text{undefined}, & x = 3 \text{ or } x = 4 \end{cases}$$ Would you like further details on any part? Here are five questions related to this topic for further exploration: 1. What would happen to the graph if the function was $$f(x) = \frac{1}{(x - 3)^3}$$? 2. How does shifting the function $$f(x) = \frac{1}{(x - 3)^2}$$ vertically affect the graph and range? 3. What is the significance of the vertical asymptote in this function's behavior? 4. How would you determine the horizontal asymptote of a rational function in general? 5. Can a rational function have more than one vertical asymptote? Why or why not? **Tip**: When dealing with rational functions, always identify the points where the denominator becomes zero to find the vertical asymptotes and undefined points.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Rational Functions
Graphing Functions
Asymptotes
Domains and Ranges
Formulas
f(x) = 1 / (x - 3)^2
Domain: x ∈ ℝ, x ≠ 3
Range: f(x) > 0
Theorems
Vertical Asymptotes occur where the denominator of a rational function is zero
Suitable Grade Level
Grades 10-12
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