The function $f(x)=x^2+3, x \geq 0$ is one-to-one.
(a) Find the inverse of $f$ and check the answer.
(b) Find the domain and the range of $f$ and $f^{-1}$.
(c) Graph $f, f^{-1}$, and $y=x$ on the same coordinate axes.
(a) $f^{-1}(x)=\sqrt{x-3}$
(Simplify your answer. Use integers or fractions for any numbers in the expression.)
(b) Find the domain of f. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The domain is \{x \mid x \leq $\square$ \}
B. The domain is \{x \mid x \geq $\square$\}
C. The domain is \{x \mid x \neq $\square$\}
D. The domain is the set of all real numbers.
Answer (2)
Verify the inverse function: f ( f − 1 ( x )) = x and f − 1 ( f ( x )) = x .
Find the domain of f ( x ) : x ≥ 0 .
Find the range of f ( x ) : f ( x ) ≥ 3 .
The domain of f is x ≥ 0 .
Explanation
Problem Analysis We are given the function f ( x ) = x 2 + 3 for = "0"> x " >= "0 . We need to find its inverse, verify the inverse, find the domain and range of both f and f − 1 , and graph them along with y = x .
Verifying the Inverse The inverse function is given as f − 1 ( x ) = x − 3 . Let's verify this by checking if f ( f − 1 ( x )) = x and f − 1 ( f ( x )) = x .
Checking f(f^{-1}(x)) First, let's compute f ( f − 1 ( x )) :
f ( f − 1 ( x )) = f ( x − 3 ) = ( x − 3 ) 2 + 3 = ( x − 3 ) + 3 = x This holds true for x ≥ 3 (since we need x − 3 ≥ 0 for the square root to be defined).
Checking f^{-1}(f(x)) Now, let's compute f − 1 ( f ( x )) :
f − 1 ( f ( x )) = f − 1 ( x 2 + 3 ) = ( x 2 + 3 ) − 3 = x 2 = ∣ x ∣ Since we are given that x ≥ 0 , we have ∣ x ∣ = x . Thus, f − 1 ( f ( x )) = x for x ≥ 0 .
Conclusion of Verification Thus, the inverse function f − 1 ( x ) = x − 3 is verified.
Domain and Range of f(x) Now, let's find the domain and range of f ( x ) = x 2 + 3 for x ≥ 0 .
Domain of f(x) The domain of f ( x ) is given as x ≥ 0 . So the domain is [ 0 , ∞ ) .
Range of f(x) To find the range of f ( x ) , we note that since x ≥ 0 , x 2 ≥ 0 . Therefore, x 2 + 3 ≥ 3 . So the range of f ( x ) is [ 3 , ∞ ) .
Domain and Range of f^{-1}(x) Now, let's find the domain and range of f − 1 ( x ) = x − 3 .
Domain of f^{-1}(x) For the domain of f − 1 ( x ) , we need x − 3 ≥ 0 , which means x ≥ 3 . So the domain of f − 1 ( x ) is [ 3 , ∞ ) .
Range of f^{-1}(x) For the range of f − 1 ( x ) , since the square root function always returns non-negative values, x − 3 ≥ 0 . So the range of f − 1 ( x ) is [ 0 , ∞ ) .
Final Answer for Domain of f The domain of f is x ≥ 0 , which corresponds to option B.
Graphing The graphs of f ( x ) , f − 1 ( x ) , and y = x can be plotted on the same coordinate axes.
Examples
Understanding inverse functions is crucial in many scientific and engineering applications. For example, if you're designing a system where you need to convert temperature from Celsius to Fahrenheit and back, the conversion formulas are inverse functions of each other. Similarly, in cryptography, encryption and decryption algorithms are inverse functions, ensuring secure communication. The ability to find and verify inverse functions allows for reversible processes, which are essential in data processing, signal analysis, and control systems.
The inverse of the function f ( x ) = x 2 + 3 for x ≥ 0 is f − 1 ( x ) = x − 3 . The domain of f is { x ∣ x ≥ 0 } , and its range is { y ∣ y ≥ 3 } . The domain and range of f − 1 are { x ∣ x ≥ 3 } and { y ∣ y ≥ 0 } respectively, and appropriate graphs can be plotted for visualization.
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