Composite functions #1 - Questions and Answers

Question 1 How do you find 𝑓 ( 𝑔   of   ℎ ) f(gofh) if 𝑓 ( 𝑥 ) = 𝑥 2 + 1 f(x)=x 2 +1, 𝑔 ( 𝑥 ) = 2 𝑥 g(x)=2x, and ℎ ( 𝑥 ) = 𝑥 − 1 h(x)=x−1? Answer Given: 𝑓 ( 𝑥 ) = 𝑥 2 + 1 f(x)=x 2 +1 𝑔 ( 𝑥 ) = 2 𝑥 g(x)=2x ℎ ( 𝑥 ) = 𝑥 − 1 h(x)=x−1 One way of thinking about these function compositions is to go back and forth between the symbols and verbal descriptions of what the functions do. In our example: 𝑓 f takes the square of a number and adds 1; 𝑔 g doubles a number; ℎ h subtracts 1 from a number. So, a verbal description of the composed 𝑓   𝑜   𝑔   𝑜   ℎ fogoh as a sequence of steps might be: Subtract 1. Double. Square. Add 1. In symbols: 𝑥 → 𝑥 − 1 → 2 ( 𝑥 − 1 ) → ( 2 ( 𝑥 − 1 ) ) 2 → ( 2 ( 𝑥 − 1 ) ) 2 + 1 x→x−1→2(x−1)→(2(x−1)) 2 →(2(x−1)) 2 +1 So: ( 𝑓   𝑜   𝑔   𝑜   ℎ ) ( 𝑥 ) = 𝑓 ( 𝑔 ( ℎ ( 𝑥 ) ) ) (fogoh)(x)=f(g(h(x))) = ( ( 2 ( 𝑥 − 1 ) ) 2 + 1 ) =((2(x−1)) 2 +1) = 4 𝑥 2 − 8 𝑥 + 5 =4x 2 −8x+5. Question 2 Use the definition of continuity and the properties of limits to show that the function is continuous at the given number. 𝑓 ( 𝑥 ) = 𝑥 2 + 7 − 𝑥 ,   𝑎 = 4 f(x)=x 2 + 7−x ​ ,a=4. Answer lim ⁡ 𝑥 → 4 𝑥 2 + lim ⁡ 𝑥 → 4 7 − 𝑥   →   Apply the Sum Law, Difference Law, and Root Law. x→4 lim ​ x 2 + x→4 lim ​ 7−x ​ →Apply the Sum Law, Difference Law, and Root Law. ( 4 2 ) + 7 − 4   →   Plug in the corresponding values. (4 2 )+ 7−4 ​ →Plug in the corresponding values. 𝑓 ( 4 ) = 16 + 3   Evaluate. f(4)=16+ 3 ​ Evaluate. The function 𝑓 ( 𝑥 ) f(x) is defined and equal to 16 + 3 16+ 3 ​ at 𝑥 = 4 x=4, thus it is continuous at that point. Question 3 Given 𝑓 ( 𝑥 ) = 4 𝑥 2 − 3   and   𝑔 ( 𝑥 ) = 6 − 1 𝑥 2 f(x)=4x 2 −3andg(x)=6− x 2 1 ​ . (b) ( 𝑔   of   𝑓 ) ( 2 ) (goff)(2) Answer Obtain the composite function ( 𝑔   𝑜   𝑓 ) ( 𝑥 ) (gof)(x). The given two functions are 𝑓 ( 𝑥 ) = 4 𝑥 2 − 3   and   𝑔 ( 𝑥 ) = 6 − 1 𝑥 2 f(x)=4x 2 −3andg(x)=6− x 2 1 ​ . To obtain the composite function ( 𝑔   𝑜   𝑓 ) ( 𝑥 ) (gof)(x), substitute 𝑓 ( 𝑥 ) f(x) for 𝑥 x in the function 𝑔 ( 𝑥 ) g(x). The composite function ( 𝑔   𝑜   𝑓 ) ( 𝑥 ) (gof)(x) is obtained as: 𝑔 ( 𝑓 ( 𝑥 ) ) = 6 − 1 ( 4 𝑥 2 − 3 ) g(f(x))=6− (4x 2 −3) 1 ​ To calculate ( 𝑔   𝑜   𝑓 ) ( 2 ) (gof)(2): Substitute 𝑥 = 2 x=2 into the composite function: 𝑔 ( 𝑓 ( 2 ) ) = 6 − 1 ( 4 ( 2 ) 2 − 3 ) g(f(2))=6− (4(2) 2 −3) 1 ​ = 6 − 1 32 − 3 = 6 − 1 29 . =6− 32−3 1 ​ =6− 29 1 ​ . Final result: ( 𝑔   𝑜   𝑓 ) ( 2 ) = 5.966 (gof)(2)=5.966. Question 4 Evaluate the function at given value of the independent variable. Simplify the results. 𝑓 ( 𝑥 ) = 𝑥 + 5 f(x)= x+5 ​ 𝑓 ( 11 ) f(11) Answer Substitute 𝑥 = 11 x=11 into 𝑓 ( 𝑥 ) = 𝑥 + 5 f(x)= x+5 ​ : 𝑓 ( 11 ) = 11 + 5 = 16 = 4. f(11)= 11+5 ​ = 16 ​ =4. Question 5 Find the composite functions 𝑓   𝑜   𝑔   and   𝑔   𝑜   𝑓 fogandgof. Find the domain of each composite function. Are the two composite functions equal? 𝑓 ( 𝑥 ) = 3 𝑥 + 1   and   𝑔 ( 𝑥 ) = − 𝑥 f(x)=3x+1andg(x)=−x. Answer Consider the function: 𝑓 ( 𝑥 ) = 3 𝑥 + 1   and   𝑔 ( 𝑥 ) = − 𝑥 f(x)=3x+1andg(x)=−x. To calculate 𝑓   𝑜   𝑔 fog: ( 𝑓   𝑜   𝑔 ) ( 𝑥 ) = 𝑓 ( 𝑔 ( 𝑥 ) ) (fog)(x)=f(g(x)) = 3 ( − 𝑥 ) + 1 = − 3 𝑥 + 1. =3(−x)+1=−3x+1. To calculate 𝑔   𝑜   𝑓 gof: ( 𝑔   𝑜   𝑓 ) ( 𝑥 ) = 𝑔 ( 𝑓 ( 𝑥 ) ) (gof)(x)=g(f(x)) = − ( 3 𝑥 + 1 ) = − 3 𝑥 − 1. =−(3x+1)=−3x−1. To calculate the domain of 𝑓   𝑜   𝑔   and   𝑔   𝑜   𝑓 fogandgof: The domain of a function is the set of input values for which the function is real or defined. Here, 𝑓   𝑜   𝑔   and   𝑔   𝑜   𝑓 fogandgof have no undefined points. Therefore, the domain of both composite functions is − ∞ < 𝑥 < ∞ −∞