1.2 Q-3

Question Statement

Given the following functions, without finding the inverse, state the domain and range of the inverse function $f^{-1}(x)$:

1. $f(x) = \sqrt{x + 2}$
2. $f(x) = \frac{x - 1}{x - 4}, x \neq 4$
3. $f(x) = \frac{1}{x + 3}, x \neq -3$
4. $f(x) = (x - 5)^2, x \geq 5$

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Background and Explanation

To solve these problems, we need to recall the relationship between a function and its inverse. Specifically:

• The domain of the function becomes the range of the inverse.
• The range of the function becomes the domain of the inverse. Additionally, we need to carefully analyze any restrictions on $x$ to identify the correct domains and ranges for both the function and its inverse.

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Solution

1. $f(x) = \sqrt{x + 2}$

• Domain of $f(x)$: For $f(x) = \sqrt{x + 2}$ to be defined, the expression under the square root must be non-negative. Therefore, $x + 2 \geq 0$, which simplifies to $x \geq -2$. Hence, the domain of $f(x)$ is $[-2, \infty)$.
• Range of $f(x)$: Since the square root function only produces non-negative values, the range of $f(x)$ is $[0, \infty)$.
• Domain and Range of $f^{-1}(x)$:
  • The domain of $f^{-1}(x)$ is the range of $f(x)$, which is $[0, \infty)$.
  • The range of $f^{-1}(x)$ is the domain of $f(x)$, which is $[-2, \infty)$.

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2. $f(x) = \frac{x - 1}{x - 4}, x \neq 4$

• Domain of $f(x)$: $f(x)$ is undefined when $x = 4$, so the domain of $f(x)$ is $(-\infty, 4) \cup (4, \infty)$.
• Range of $f(x)$: Since the function is a rational expression with a vertical asymptote at $x = 4$, its range is $(-\infty, 4) \cup (4, \infty)$.
• Domain and Range of $f^{-1}(x)$:
  • The domain of $f^{-1}(x)$ is the range of $f(x)$, which is $(-\infty, 4) \cup (4, \infty)$.
  • The range of $f^{-1}(x)$ is the domain of $f(x)$, which is $(-\infty, 4) \cup (4, \infty)$.

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3. $f(x) = \frac{1}{x + 3}, x \neq -3$

• Domain of $f(x)$: The function is undefined when $x = -3$, so the domain of $f(x)$ is $(-\infty, -3) \cup (-3, \infty)$.
• Range of $f(x)$: Since the function is a rational expression with no restrictions other than $x \neq -3$, the range of $f(x)$ is $(-\infty, 0) \cup (0, \infty)$.
• Domain and Range of $f^{-1}(x)$:
  • The domain of $f^{-1}(x)$ is the range of $f(x)$, which is $(-\infty, 0) \cup (0, \infty)$.
  • The range of $f^{-1}(x)$ is the domain of $f(x)$, which is $(-\infty, -3) \cup (-3, \infty)$.

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4. $f(x) = (x - 5)^2, x \geq 5$

• Domain of $f(x)$: The function is defined for $x \geq 5$, so the domain of $f(x)$ is $[5, \infty)$.
• Range of $f(x)$: The expression $(x - 5)^2$ produces only non-negative values, so the range of $f(x)$ is $[0, \infty)$.
• Domain and Range of $f^{-1}(x)$:
  • The domain of $f^{-1}(x)$ is the range of $f(x)$, which is $[0, \infty)$.
  • The range of $f^{-1}(x)$ is the domain of $f(x)$, which is $[5, \infty)$.

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Key Formulas or Methods Used

• Domain and Range of Inverse Function:
  • Domain of $f^{-1}(x)$ = Range of $f(x)$
  • Range of $f^{-1}(x)$ = Domain of $f(x)$

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Summary of Steps

1. Identify the domain of the given function $f(x)$ based on its definition and any restrictions.
2. Determine the range of the function by analyzing how the function behaves over its domain.
3. Use the inverse function properties: The domain of $f^{-1}(x)$ is the range of $f(x)$, and the range of $f^{-1}(x)$ is the domain of $f(x)$.
4. State the domain and range of the inverse function based on the relationships above.