13. Find the domain of [tex]f(x)=\sqrt{4-x^2}[/tex].
A. [tex](-2,2)[/tex]
B. [tex](-2,2)[/tex]
C. [tex](-2,2)[/tex]
D. [tex](-2,2)[/tex].
14. Find the domain of [tex]f(x)=\frac{5}{x^2-9}[/tex].
A. [tex]R \{9\}[/tex]
B. [tex]R \{-3,3\}[/tex]
C. [tex]R \{3\}[/tex]
D. [tex]R \{-9\}[/tex]
15. The domain of [tex]f(x)=x^2 \ln x[/tex] is
A. [tex](-\infty, x)[/tex]
B. [tex](1, \infty)[/tex]
C. [tex](0,1)[/tex]
D. [tex](0, x)[/tex].
16. Find the domain of [tex]\sqrt{4-7 t}[/tex].
A. [tex][0, \infty)[/tex]
B. [tex]\frac{7}{1}[/tex]
C. [tex](-\infty, \frac{7}{1}][/tex]
D. [tex][\frac{7}{1}, \infty)[/tex].
17. Let [tex]f: R \rightarrow R[/tex] be defined by [tex]f(x)=\frac{5}{E^2-1}[/tex]. Find the domain of [tex]f[/tex].
A. [tex]R \{4\}[/tex]
B. [tex]R \{2\}[/tex]
C. [tex]R \{-4\}[/tex]
D. [tex]R \{-2,2\}[/tex].
18. Find the inverse [tex]g^{-1}(x)[/tex] of [tex]g(x)=\sqrt{x-3}[/tex]
A. [tex](x-3)^2[/tex]
B. [tex]x^2+3[/tex]
C. [tex]x^2-3[/tex]
Answer (2)
The domain of f ( x ) = 4 − x 2 is [ − 2 , 2 ] .
The domain of f ( x ) = x 2 − 9 5 is R \ { − 3 , 3 } .
The domain of f ( x ) = x 2 ln x is ( 0 , ∞ ) .
The domain of 4 − 7 t is ( − ∞ , 7 4 ] .
The domain of f ( x ) = x 2 − 1 5 is R \ { − 1 , 1 } .
The inverse of g ( x ) = x − 3 is x 2 + 3 .
Explanation
Introduction We will solve each question step by step, finding the domain of the given functions and the inverse of a given function.
Question 13 Solution Question 13: Find the domain of f ( x ) = 4 − x 2 . The domain is found by solving the inequality 4 − x 2 ≥ 0 . This is equivalent to x 2 ≤ 4 , which means − 2 ≤ x ≤ 2 . Therefore, the domain is [ − 2 , 2 ] .
Question 14 Solution Question 14: Find the domain of f ( x ) = x 2 − 9 5 . The domain is found by solving x 2 − 9 n e 0 . This is equivalent to x 2 n e 9 , which means x n e − 3 and x n e 3 . Therefore, the domain is R \ { − 3 , 3 } .
Question 15 Solution Question 15: The domain of f ( x ) = x 2 ln x is found by solving 0"> x > 0 since the natural logarithm is only defined for positive numbers. Therefore, the domain is ( 0 , ∞ ) .
Question 16 Solution Question 16: Find the domain of 4 − 7 t . The domain is found by solving the inequality 4 − 7 t ≥ 0 . This is equivalent to 7 t ≤ 4 , which means t ≤ 7 4 . Therefore, the domain is ( − ∞ , 7 4 ] .
Question 17 Solution Question 17: Let f : R → R be defined by f ( x ) = x 2 − 1 5 . Find the domain of f . The domain is found by solving x 2 − 1 n e 0 . This is equivalent to x 2 n e 1 , which means x n e − 1 and x n e 1 . Therefore, the domain is R \ { − 1 , 1 } .
Question 18 Solution Question 18: Find the inverse g − 1 ( x ) of g ( x ) = x − 3 . To find the inverse, set y = x − 3 . Then, square both sides to get y 2 = x − 3 . Solve for x to get x = y 2 + 3 . Therefore, g − 1 ( x ) = x 2 + 3 .
Final Answers Final Answers:
The domain of f ( x ) = 4 − x 2 is [ − 2 , 2 ] .
The domain of f ( x ) = x 2 − 9 5 is R \ { − 3 , 3 } .
The domain of f ( x ) = x 2 ln x is ( 0 , ∞ ) .
The domain of 4 − 7 t is ( − ∞ , 7 4 ] .
The domain of f ( x ) = x 2 − 1 5 is R \ { − 1 , 1 } .
The inverse of g ( x ) = x − 3 is x 2 + 3 .
Examples
Understanding domains and inverses is crucial in many real-world applications. For example, when designing a bridge, engineers need to consider the domain of the load-bearing function to ensure the structure's stability under various weights. Similarly, in cryptography, understanding the inverse of an encryption function is essential for decrypting messages and maintaining secure communications. These concepts provide a foundation for analyzing and solving complex problems in engineering, computer science, and beyond.
The domains for the functions in the questions are as follows: 13: [ − 2 , 2 ] ; 14: R ∖ { − 3 , 3 } ; 15: ( 0 , ∞ ) ; 16: ( − ∞ , 7 4 ] ; 17: R ∖ { − 1 , 1 } ; 18: the inverse is g − 1 ( x ) = x 2 + 3 .
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