8. Given [tex]g(x)=x^3[/tex] and [tex]h(x)=x+1[/tex], find [tex](g \circ h)(-2)[/tex].
A. 1
B. -1
C. 2
D. -2
9. If [tex]f(x)=x^2-1[/tex] and [tex]g(x)=\sqrt{x}+1[/tex]. Find [tex](f \circ g)(x)[/tex]
A. [tex]x+2 \sqrt{x}[/tex]
B. [tex]x^2[/tex]
C. 2
D. [tex]\sqrt{x}-1[/tex].
10. If [tex]f(x)=10 x-5[/tex] and [tex]g(x)=x+3[/tex], find [tex](f \circ g)(x)[/tex].
A. [tex]10 x+25[/tex]
B. [tex]20 x-5[/tex]
C. [tex]10 x-25[/tex]
D. [tex]2 x-25[/tex].
11. Given [tex]f(x)=3 x^2[/tex] and [tex]g(x)=1+2 x[/tex] with [tex]x \in R[/tex], then [tex](f \circ g)(0)[/tex] is
A. 3
B. 0
C. 1
D. 2.
12. If [tex]f: X \rightarrow Y[/tex] is such that [tex]f(X)=Y[/tex], then [tex]f[/tex] is ?
A. undefine
B. onto
C. bijective
D. one-to-one
Answer (2)
The answers to the questions are as follows: 8) -1 (Option B), 9) x + 2√x (Option A), 10) 10x + 25 (Option A), 11) 3 (Option A), 12) onto (Option B).
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Question 8: Evaluate h ( − 2 ) = − 2 + 1 = − 1 , then g ( − 1 ) = ( − 1 ) 3 = − 1 . The answer is − 1 .
Question 9: Find f ( g ( x )) = f ( x + 1 ) = ( x + 1 ) 2 − 1 = x + 2 x + 1 − 1 = x + 2 x . The answer is x + 2 x .
Question 10: Find f ( g ( x )) = f ( x + 3 ) = 10 ( x + 3 ) − 5 = 10 x + 30 − 5 = 10 x + 25 . The answer is 10 x + 25 .
Question 11: Evaluate g ( 0 ) = 1 + 2 ( 0 ) = 1 , then f ( 1 ) = 3 ( 1 ) 2 = 3 . The answer is 3 .
Question 12: Since f ( X ) = Y , the function f is onto. The answer is $\boxed{onto}.
Explanation
Find h(-2) We are given g ( x ) = x 3 and h ( x ) = x + 1 , and we need to find ( g ∘ h ) ( − 2 ) . This means we need to evaluate g ( h ( − 2 )) . First, let's find h ( − 2 ) .
Calculate g(h(-2)) h ( − 2 ) = − 2 + 1 = − 1 . Now we need to find g ( h ( − 2 )) , which is g ( − 1 ) .
Final Answer for Question 8 g ( − 1 ) = ( − 1 ) 3 = − 1 . Therefore, ( g ∘ h ) ( − 2 ) = − 1 .
Find f(g(x)) We are given f ( x ) = x 2 − 1 and g ( x ) = x + 1 , and we need to find ( f ∘ g ) ( x ) . This means we need to evaluate f ( g ( x )) . We substitute g ( x ) into f ( x ) .
Expand the expression f ( g ( x )) = f ( x + 1 ) = ( x + 1 ) 2 − 1 . Expanding the square, we get ( x + 1 ) 2 = ( x ) 2 + 2 ( x ) ( 1 ) + 1 2 = x + 2 x + 1 .
Final Answer for Question 9 So, f ( g ( x )) = x + 2 x + 1 − 1 = x + 2 x . Therefore, ( f ∘ g ) ( x ) = x + 2 x .
Find f(g(x)) We are given f ( x ) = 10 x − 5 and g ( x ) = x + 3 , and we need to find ( f ∘ g ) ( x ) . This means we need to evaluate f ( g ( x )) . We substitute g ( x ) into f ( x ) .
Final Answer for Question 10 f ( g ( x )) = f ( x + 3 ) = 10 ( x + 3 ) − 5 = 10 x + 30 − 5 = 10 x + 25 . Therefore, ( f ∘ g ) ( x ) = 10 x + 25 .
Find g(0) We are given f ( x ) = 3 x 2 and g ( x ) = 1 + 2 x , and we need to find ( f ∘ g ) ( 0 ) . This means we need to evaluate f ( g ( 0 )) . First, let's find g ( 0 ) .
Calculate f(g(0)) g ( 0 ) = 1 + 2 ( 0 ) = 1 . Now we need to find f ( g ( 0 )) , which is f ( 1 ) .
Final Answer for Question 11 f ( 1 ) = 3 ( 1 ) 2 = 3 ( 1 ) = 3 . Therefore, ( f ∘ g ) ( 0 ) = 3 .
Determine the type of function We are given that f : X → Y is such that f ( X ) = Y , and we need to determine what type of function f is.
Final Answer for Question 12 Since f ( X ) = Y , this means that the image of the set X under the function f is equal to the set Y . In other words, every element in Y has a pre-image in X . This is the definition of an onto function (also known as a surjective function).
Examples
Composite functions are used in many real-world applications. For example, in economics, the cost of producing x items might be a function C ( x ) , and the number of items produced as a function of time t might be x ( t ) . Then the cost of production as a function of time would be the composite function C ( x ( t )) . Similarly, in physics, the position of an object might be a function of time p ( t ) , and the velocity of the object might be a function of its position v ( p ) . Then the velocity as a function of time would be the composite function v ( p ( t )) .
