Questions and Answers #2 Composite Functions

Questions and Answers Sheet 2 Composite Functions Question #1 What is the continuity of the composite function f(g(x)) given 𝑓(𝑥) = 1 √𝑥 and 𝑔(𝑥) = 𝑥 − 1 Answer: Explanation: The process here is to verify first the domain of g(x) (the inner function in the composition). The domain is {𝑥 ∣ 𝑥 > 1, 𝑥 ∈ 𝑅𝑅} 1 𝑓(𝑔(𝑥)) = √𝑥 There will be two types of restriction on the domain in this problem. - When the number underneath the √ is less than 0. - When the denominator equals 0. The number under the square root will be negative whenever x<1 . The denominator will equal 0 when x=1 , so the domain of the composition is {𝑥 ∣ 𝑥 > 1, 𝑥 ∈ 𝑅𝑅 } Question #2 How do you write 𝑗(𝑥) = sin2 (𝑥) as a composition of two or more functions? Answer: Let 𝑓(𝑥) = sin 𝑥 and 𝑔(𝑥) = 𝑥 2 Then 𝑗(𝑥) = (𝑔 ∘ 𝑓)(𝑥) = 𝑔[𝑓(𝑥)] = 𝑔(sin 𝑥) = sin2 𝑥 Question #3 Let F(x)=3x and 𝑔(𝑦) = 1/𝑦, how do you find each of the compositions and domain and range? Answer: For: (𝑓 ∘ 𝑔)(𝑥) ℎ(𝑥) = [3𝑥]𝑥=𝑔(𝑥)=1/𝑥 = 3 ∗ 1/𝑥 = 3/𝑥 For: (𝑔 ∘ 𝑓)(𝑥) 𝑟(𝑥) = [1/𝑥]𝑥=𝑓(𝑥)=3𝑥 = 1/(3𝑥) Question #4 How do you write the combined function as a composition of several functions if 𝑓(𝑔(𝑥)) = √1 − 𝑥 2 + 2 Answer: Your equation for g could possibly be √1 − 𝑥 2 After, we plug in x into that equation we get the same function (√1 − 𝑥 2 ) Next step when solving multiple step functions is to plug in your result of your first function (in our case g) into the new function (f). This is where we need another function to get √1 − 𝑥 2 into √1 − 𝑥 2 + 2 For this step we may say (𝑓) = 𝑥 + 2 When we plug in √1 − 𝑥 2 into our new (f)=x+2 our result is √1 − 𝑥 2 + 2, the equation we needed to get to. Question #5 How do you find the composition of function given 𝑓(𝑥) = √𝑥 + 8 and 𝑔(𝑥) = 4𝑥 + 1 Answer: To complete a function composition, you must substitute one function in the place of x in the other. For example in f(g(x)) , you must plug function g into f. 𝑓(𝑔(𝑥)) = √(4𝑥 + 1) + 8 𝑓(𝑔(𝑥)) = √(4𝑥 + 9) When finding the composition of functions think of working from the inside to the outside. So, if you have f(g(h(x))), you plug h into g and then you plug the result from that calculation. When there is a number in the place of x, you must plug that number in for x and then work your way out. Question #6 Let 𝑓(𝑥) = 7𝑥 2 + 5 and 𝑔(𝑥) = 𝑥 − 3 , how do you find the composite function (𝑓 ∘ 𝑔)(2)? Answer: first you have to find what f(x) is in order to get g(x) so in order to get that, you have to first find what f(2) is and you do it like this: 𝑓(𝑥) = 7(2)2 + 5 the 2 replaces x 33 and now you have to find what g(x) is so you have to take 33 and you get 𝑔(𝑥) = 33 − 3 and you get your answer Question #7 Function f and g are defined by 𝑓(𝑥) = √((𝑥 2 − 2𝑥)) and 𝑔(𝑥) = 3𝑥 + 4 . The composite function is undefined for 𝑥 ∈]𝑎; 𝑏[ . Find the value of a and b? Answer: The functions are 𝑓(𝑥) = √(𝑥 2 − 2𝑥) 𝑔(𝑥) = 3𝑥 + 4 The composite function is 𝑓(𝑔(𝑥)) = 𝑓(3𝑥 + 4) = √((3𝑥 + 4)2 − 2(3𝑥 + 4)) = √(9𝑥 2 + 24𝑥 + 16 − 6𝑥 − 8) = √(9𝑥 2 + 18𝑥 + 8) The domain is 9𝑥 2 + 18𝑥 + 8 ≥ 0 Therefore, The solution to this quadratic equation is 𝑥 = (−18 ± √(182 − 4 ∗ 9 ∗ 8)) /(18) = (−18 ± √(182 − 4 ∗ 9 ∗ 8)) /(18) = (−18 ± √36)/18 = (−18 ± 6)/18 Therefore, the composite function is undefined for (−24/18, −12/18) = (−4/3, −2/3) Question #8 How do you write 𝑘(𝑥) = 𝑒 𝑠𝑖𝑛 𝑥 as a composition of two or more functions? Answer: In precalculus-related courses, when we write f(g(x)) , the notation means to first apply the function gto the number x to get the number g(x) , then apply the function f to the number g(x) to get f(g(x)) (so we work "from inside out"). (In some other math courses, the inputs and outputs don't have to be numbers). For the function 𝑘(𝑥) = 𝑒 𝑠𝑖𝑛 𝑥 when a number x is substituted, the sine function must be evaluated first to get the number sin(x) . This number then must be made the power of e to get the final output 𝑒 𝑠𝑖𝑛 𝑥. Therefore, the sine function must be the "inner" (first applied) function and the exponential function must be the "outer" (second applied) function. Hence, k(x)=f(g(x)) when 𝑓(𝑥) = 𝑒 (𝑥) and g(x)=sin(x) Question #9 How do you find (f of g of h) if 𝑓(𝑥) = 𝑥 2 + 1 and h(x)=x-1 ? Answer: Given: (𝑓(𝑥) = 𝑥 2 + 1), (𝑔(𝑥) = 2𝑥), (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 takes the square of a number and adds 1; - g doubles a number; - h subtracts 1 from a number. So a verbal description of the composed f@g@h as a sequence of steps might be: - Subtract 1; - Double; - Square; - Add 1. So in symbols we might describe this process thus: 2 2 𝑥 → 𝑥 − 1 → 2(𝑥 − 1) → (2(𝑥 − 1)) → (2(𝑥 − 1)) + 1 So: (f@g@h)(x) = f(g(h(x))) 2 = (2(𝑥 − 1)) + 1 = 4(𝑥 2 − 2𝑥 + 1) + 1 = 4𝑥 2 − 8𝑥 + 4 + 1 = 4𝑥 2 − 8𝑥 + 5 Question #10 Suppose that f(x)=-4-3x and 𝑔(𝑥) = √2 + 𝑥 find the rule of the composite function 𝑓 ∘ 𝑔 ? Answer: 𝑓@𝑔 = 𝑓[𝑔(𝑥)] In this question, you can sub the inner part g(x) first and sub the things we get to replace the x of f(x): 𝑓@𝑔 = 𝑓[𝑔(𝑥)] = 𝑓(√2 + 𝑥) = −4 − 3(√2 + 𝑥) = −4 − 3√2 − 3𝑥 Question #11 How do you find the (𝑓 ∘ 𝑔 ∘ ℎ)(𝑥) for 𝑓(𝑥) = (𝑥 − 2)/(2𝑥 + 1), 𝑔(𝑥) = 3𝑥 + 1, ℎ(𝑥) = 𝑥 2 Answer: Start out by plugging h into g. 𝑓(𝑔(𝑥 2 )) = 𝑓(3(𝑥 2 ) + 1) = 𝑓(3𝑥 2 + 1) Next, plug in the new function into f. = (3𝑥 2 + 1 − 2)/(2(3𝑥 2 + 1) + 1) = (3𝑥 2 − 1)/(6𝑥 2 + 3) Question #12 How do you write 𝑦 = (𝑥 − 11)5 as a composition of two simpler functions? Answer: I think that writing in words what the function does is 90% of the job done. This function takes a number x and: 1) Subtract 11 from said number; 2) Computes the fifth power of the result. As you can see, we have actually written the two simplier functions: 𝑓(𝑥) = 𝑥 5 ("computes the fifth power"), and g(x)=x-11 ("subtract eleven from x") In formulas, (𝑥 − 11)5 = 𝑓(𝑔(𝑥)) Question #13 If 𝑓(𝑥) = 2𝑥 − 9 and 𝑔(𝑥) = (𝑥 + 1)2 , what is the composite function f(g(x)) Answer: Start by putting g(x) into that: 𝑓(𝑔(𝑥)) = 𝑓((𝑥 + 1)2 ) Now, if f(x) = 2x - 9 , then anything can fit inside x and be put into the equation. Let's set variable 𝑘 = (𝑥 + 1)2 and fit it inside f(x) : 𝑓((𝑥 + 1)2 ) = 𝑓(𝑘) = 2𝑘 − 9 Now we just replace k with (𝑥 + 1)2 𝑓(𝑘) = 2𝑘 − 9 = 2(𝑥 + 1)2 − 9 Since essentially 𝑘 = (𝑥 + 1)2 = 𝑔(𝑥), we now have 𝑓(𝑘) = 𝑓(𝑔(𝑥)) = 2(𝑥 + 1)2 − 9 If asked to expand, do so: 𝑓(𝑔(𝑥)) = 2(𝑥 + 1)2 − 9 = 2(𝑥 + 1)(𝑥 + 1) − 9 = 2(𝑥 2 + 𝑥 + 𝑥 + 1) − 9 = 2(𝑥 2 + 2𝑥 + 1) − 9 = 2𝑥 2 + 4𝑥 + 2 − 9 = 2𝑥 2 + 4𝑥 − 7 So we have: 𝑓(𝑔(𝑥)) = 2(𝑥 + 1)2 − 9 = 2𝑥 2 + 4𝑥 − 7 Question #14 How do you write 𝑦 = 1/(𝑥 2 + 3) as a composition of two simpler functions? Answer: You can write it as a composition of the following functions f(x)=1/x and 𝑔(𝑥) = 𝑥 2 + 3 hence the composition gives 𝑓(𝑔(𝑥)) = 1/(𝑥 2 + 3) Question #15 How do you write 𝑦 = 3√(1 + 𝑥 2 ) as a composition of two simpler functions? Answer: Let g(x) be the first thing we do if we knew x and started to calculate: 𝑔(𝑥) = 𝑥 2 Now f will be the rest of the calculation we would do (after we found 𝑥 2 ) It may be easier to think about if we gave g(x) a temporary name, say g(x)=u So we see that 𝑦 = 3√(1 + 𝑢) So 𝑓(𝑢) = 3√(1 + 𝑢) and that tells us we want: 𝑓(𝑥) = 3√(1 + 𝑥) Another answer is to letf (x) be the last thing we would do in calculating y . So let f(x) = 3x To get y = f(g(x)) we need 3g(x) = y So let 𝑔(𝑥) = √(1 + 𝑥 2 ) Question #16 How do you find the compositions given 𝑓(𝑥) = 𝑥 3 + 2 and Answer: fg(x)=x and gf(x)=x 3 x −3 ?