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, \infty)[/tex] B. [tex](1, \infty)[/tex] C. [tex](0,1)[/tex] D. [tex](0, \infty)[/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}{4}][/tex] D. [tex][\frac{7}{4}, \infty)[/tex].
17. Let [tex]f: R \rightarrow R[/tex] be defined by [tex]f(x)=\frac{5}{x^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] D. [tex]\frac{1}{x-3}[/tex].
19. The inverse of [tex]f(x)=2 x-1[/tex] is
A. [tex]x-1[/tex] B. [tex]\frac{x+1}{2}[/tex] C. [tex]\frac{1}{x+2}[/tex] D. [tex]x^2[/tex].
20. Find the inverse of [tex]f(x)=x^3-2[/tex].
A. [tex](x+2)^{\frac{1}{3}}[/tex] B. [tex](x-2)^{\frac{1}{3}}[/tex] C. [tex](x-2)[/tex] D. [tex](x+2)[/tex].
21. If [tex]h(x)=\frac{x+1}{2 x-5}[/tex], find [tex]h^{-1}(0)[/tex].
A. 0 B. -1 C. 1 D. [tex]\frac{5}{2}[/tex].

The chosen answers are as follows: 13. D. [-2,2], 14. A. R \ {-3,3}, 15. D. (0, ∞), 16. C. (-∞, 4/7], 17. None, 18. B. x^2+3, 19. B. \frac{x+1}{2}, 20. A. (x+2)^{1/3}, 21. B. -1.
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Answered by Anonymous | 2025-07-07

Domain of 4 − x 2 ​ : [ − 2 , 2 ] .
Domain of x 2 − 9 5 ​ : R \ { − 3 , 3 } .
Domain of x 2 ln x : ( 0 , ∞ ) .
Domain of 4 − 7 t ​ : ( − ∞ , 7 4 ​ ] .
Domain of x 2 − 1 5 ​ : R \ { − 1 , 1 } .
Inverse of x − 3 ​ : x 2 + 3 .
Inverse of 2 x − 1 : 2 x + 1 ​ .
Inverse of x 3 − 2 : ( x + 2 ) 3 1 ​ .
h − 1 ( 0 ) for h ( x ) = 2 x − 5 x + 1 ​ : -1. − 1 ​

Explanation

Introduction We will solve questions 13-21, which involve finding domains and inverses of functions.

Question 13 For question 13, we need to find the domain of f ( x ) = 4 − x 2 ​ . The expression inside the square root must be non-negative, so we need to solve 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 For question 14, we need to find the domain of f ( x ) = x 2 − 9 5 ​ . The denominator cannot be zero, so we need to find the values of x for which x 2 − 9 = 0 . This gives x 2 = 9 , so x = ± 3 . Therefore, the domain is all real numbers except x = ± 3 , which can be written as R \ { − 3 , 3 } .

Question 15 For question 15, we need to find the domain of f ( x ) = x 2 ln x . The natural logarithm function ln x is only defined for 0"> x > 0 . Therefore, the domain of f ( x ) is ( 0 , ∞ ) .

Question 16 For question 16, we need to find the domain of 4 − 7 t ​ . The expression inside the square root must be non-negative, so we need to solve the inequality 4 − 7 t ≥ 0 . This gives 7 t ≤ 4 , so t ≤ 7 4 ​ . Therefore, the domain is ( − ∞ , 7 4 ​ ] .

Question 17 For question 17, we need to find the domain of f ( x ) = x 2 − 1 5 ​ . The denominator cannot be zero, so we need to find the values of x for which x 2 − 1 = 0 . This gives x 2 = 1 , so x = ± 1 . Therefore, the domain is all real numbers except x = ± 1 , which can be written as R \ { − 1 , 1 } .

Question 18 For question 18, we need to find the inverse of g ( x ) = x − 3 ​ . To find the inverse, we set y = x − 3 ​ . Then y 2 = x − 3 , so x = y 2 + 3 . Therefore, g − 1 ( x ) = x 2 + 3 . The domain of g ( x ) is x ≥ 3 , and the range is y ≥ 0 . The domain of g − 1 ( x ) is x ≥ 0 .

Question 19 For question 19, we need to find the inverse of f ( x ) = 2 x − 1 . To find the inverse, we set y = 2 x − 1 . Then 2 x = y + 1 , so x = 2 y + 1 ​ . Therefore, f − 1 ( x ) = 2 x + 1 ​ .

Question 20 For question 20, we need to find the inverse of f ( x ) = x 3 − 2 . To find the inverse, we set y = x 3 − 2 . Then x 3 = y + 2 , so x = ( y + 2 ) 3 1 ​ . Therefore, f − 1 ( x ) = ( x + 2 ) 3 1 ​ .

Question 21 For question 21, we are given h ( x ) = 2 x − 5 x + 1 ​ , and we need to find h − 1 ( 0 ) . This means we need to find the value of x such that h ( x ) = 0 . So, 2 x − 5 x + 1 ​ = 0 . This implies x + 1 = 0 , so x = − 1 . Therefore, h − 1 ( 0 ) = − 1 .

Final Answers In summary:

D. [ − 2 , 2 ]

B. R \ { − 3 , 3 }

D. ( 0 , ∞ )

C. ( − ∞ , 7 4 ​ ]

None of the options are correct. The correct answer is R \ { − 1 , 1 } .

B. x 2 + 3

B. 2 x + 1 ​

A. ( x + 2 ) 3 1 ​

B. -1


Examples
Understanding domains and inverses of functions is crucial in many real-world applications. For example, in physics, if you have a function that describes the position of an object over time, finding the domain tells you the valid time intervals for which the position is defined. Finding the inverse allows you to determine the time at which the object was at a specific position. These concepts are also fundamental in engineering, computer science, and economics, where functions are used to model various systems and processes.

Answered by GinnyAnswer | 2025-07-07