Topic 6 - Square Root Functions & Equations 6-1 Square Root Functions as Inverses 6-2 Attributes of Square Root Functions 6-3 Transformations of Square Root Functions 6-4 Introduction to Square Root Equations 6-5 Solving Square Root Equations

6-1 Square Root Functions as Inverses Objectives: Describe and analyze the relationship between a quadratic function and its square root inverse. Use composition of quadratic and square root functions to determine if they are inverses.

one-to-one function inverse relation 𝑓 −1 inverse functions 6-1 Square Root Functions as Inverses - Additional Vocabulary Support Choose the word or phrase from the list that best matches each sentence. one-to-one function inverse relation 𝑓 −1 inverse functions one-to-one function In a(n) , each y-value in the range corresponds to exactly one x-value in the domain. A relation pairs element a of its domain to element b of its range. The pairs b with a. 3. The inverse of a function f is represented by . If a relation and its inverse are functions, then they are . inverse relation 𝑓 −1 inverse functions

6-1 Square Root Functions as Inverses - Additional Vocabulary Support

6-1 Square Root Functions as Inverses - Additional Vocabulary Support Replace f (x) with y. Switch x and y. Square both sides. Add 2 to both sides. Divide both sides by 3 to solve for y.

6-1 Square Root Functions as Inverses - Additional Vocabulary Support Replace f (x) with y. Switch x and y. Add 8 to both sides Divide both sides by 4 to solve for y. .

6-1 Square Functions as Inverses Inverse of a Quadratic Function A horizontal line can intersect the graph of 𝑓 𝑥 = 𝑥 2 in two points, i.e.,f (-2) = f (2). Thus, a vertical line can intersect the graph of 𝑓 −1 in two points. 𝑓 −1 is not a function because it fails the vertical line test. Restrict the domain of f so that the inverse of the restricted function is a function.

6-1 Square Functions as Inverses A landscaper is designing a square garden and wants to put edging along one side. The edging costs $1 per linear foot. As part of a cost analysis, the landscaper examines the cost of the edging as a function of the area of the garden. Graph this relationship. What is an equation of this graph? Explain your reasoning. Let A = area of garden 𝑪= 𝑨 C = cost Cost, C Area, A Cost (C) 1 1 2 4 3 9 5 25 Area (A)

6-1 Square Functions as Inverses Finding an Inverse Consider the function 𝒚= 𝒙 𝟐 +𝟓. a. Write the inverse of the function. 𝒚=± 𝒙−𝟓 b. Analyze and describe the relationship between the function and its inverse, including restrictions on domain and range. The given function is quadratic so its inverse is square root. Both are graphed as parabolas. The domain 𝐨𝐟 the function is  −∞, ∞ and range is 𝒚≥𝟓. The domain 𝐨𝐟 the inverse is 𝒙≥𝟓  and range is −∞, ∞ .

6-1 Square Functions as Inverses c. Graph the inverse.

6-1 Square Functions as Inverses Graphing a Function and Its Inverse Consider the function 𝒇 𝒙 =𝟒 𝒙 𝟐 −𝟏 with domain 𝒙>𝟎, and its inverse, 𝒇 −𝟏 𝒙 = 𝒙+𝟏 𝟐 . Analyze and describe the relationship between the function and its inverse. The function 𝒇 𝒙 is half of a quadratic function, and its inverse is a square root. The graph of 𝒇 𝒙 opens upward and the graph of the inverse is reflected on the line 𝒚=𝒙, so it opens to the right.

6-1 Square Functions as Inverses Graphing a Function and Its Inverse Consider the function 𝒇 𝒙 =𝟒 𝒙 𝟐 −𝟏 with domain 𝒙>𝟎, and its inverse, 𝒇 −𝟏 𝒙 = 𝒙+𝟏 𝟐 . c. Graph f (x) and f -1(x). b. What are the domain and range of f (x) and f -1(x)? Write your answers as inequalities. For 𝒇 𝒙 , domain is 𝒙>𝟎, and its range is 𝒇 𝒙 >−𝟏. For 𝒇 −𝟏 𝒙 , domain is 𝒙>−𝟏, and its range is 𝒇 −𝟏 𝒙 > 0.

6-1 Square Functions as Inverses Finding the Inverse of a Formula The function 𝑑= 𝑣 2 19.6 relates the distance d, in meters, that an object has fallen to its velocity v, in meters per second. Find the inverse of this function. What is the velocity of a cliff diver in meters per second as he enters the water, if he dives from 24 meters above the water? SOLUTION: 𝒅= 𝒗 𝟐 𝟏𝟗.𝟔  𝟏𝟗.𝟔𝒅= 𝒗 𝟐  𝒗= 𝟏𝟗.𝟔𝒅 𝒗= 𝟏𝟗.𝟔 𝟐𝟒 Solve for 𝒗 when 𝒅=𝟐𝟒 𝒎:  𝒗≈𝟐𝟏.𝟕 𝒎/𝒔

6-1 Square Functions as Inverses Composing Functions Use composition to show that 𝒇 𝒙 = 𝒙−𝟏 𝟐 with domain 𝒙≥𝟏, and 𝒈 𝒙 = 𝒙 +𝟏 are inverse functions. Recall. 𝑓 and 𝑓 −1 are inverse functions if and only if 𝒇∘ 𝒇 −𝟏 𝒙 =𝒙 and 𝒇 −𝟏 ∘𝒇 𝒙 =𝒙 for 𝒙 in the domains of 𝒇 and 𝒇 −𝟏 , respectively. Solution: We need to show 𝒇∘𝒈 = 𝒈∘𝒇 =𝒙. = 𝒇 𝒈 𝒙 𝒇∘𝒈 = 𝒈 𝒇 𝒙 𝒈∘𝒇 = 𝒇 𝒙 +𝟏 = 𝒙 +𝟏 −𝟏 𝟐 = 𝒈 𝒙−𝟏 𝟐 = 𝒙−𝟏 𝟐 +𝟏 = 𝒙 +𝟏 𝟐 −𝟐 𝟏 𝒙 +𝟏 + 𝟏 𝟐 = (𝒙+𝟐 𝒙 + 1)−𝟐 𝒙 −𝟐+𝟏 = 𝒙−𝟏 +𝟏 = 𝒙+𝟐 𝒙 −𝟐 𝒙 −𝟐+𝟏+ 1 𝒈∘𝒇 =𝒙 𝒇∘𝒈 =𝒙

6-1 Square Functions as Inverses 1. Find the inverse of 𝒇 𝒙 = 𝒙+𝟒 𝟐 and sketch its graph. Is the inverse a function? 𝒇 −𝟏 𝒙 =± 𝒙 −𝟒 ; 𝒇 −𝟏 𝐢𝐬 𝐧𝐨𝐭 𝐚 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧.

6-1 Square Functions as Inverses 2. Explain how you can find the domain and range of the inverse of 𝒇 𝒙 = 𝒙+𝟗 without first finding the equation of the inverse. What are the domain and range of the inverse? The range of 𝒇 −𝟏 𝒙 is the domain of 𝒇 𝒙 , and the domain of 𝒇 −𝟏 𝒙 is the range of 𝒇 𝒙 , Domain of 𝒇 −𝟏 𝒙 : 𝒙≥𝟎 Range of 𝒇 −𝟏 𝒙 : 𝒚≥−𝟗

6-1 Square Functions as Inverses 3. The height in feet of a football when it has travelled a horizontal distance of x feet from the spot where it was kicked is modeled by the function 𝒇 𝒙 =−𝟒 𝒙−𝟒 𝟐 +𝟔𝟎. Does 𝒇 −𝟏 𝟕𝟓 exist? If so, calculate it. If not, explain what that tells you about the path of the football. 𝒇 −𝟏 𝟕𝟓 does not exist; the football never reaches a height of 75 feet.

6-1 Square Functions as Inverses 𝑔 𝑥 = 𝑥 2 4. Vocabulary Does every square root function have an inverse that is a function? Does every quadratic function have an inverse that is a function? YES NO 5. Analyze Mathematical Relationships (1)(F) Define 𝑔 𝑥 = 𝑥 2 and define f (x) as follows: 𝑓 𝑥 = 𝑥 2 for 𝑥≥0, and 𝑓 𝑥 is not defined for 𝑥<0. Explain why 𝑓 −1 𝑥 is a function but 𝑔 −1 𝑥 is not. 𝒇 −𝟏 𝒙 passes the vertical line test but 𝒈 −𝟏 𝒙 does not. 𝒚=𝒙 𝒈 𝒙 = 𝒙 𝟐 𝒚=𝒙 𝒇 𝒙 = 𝒙 𝟐 , 𝒙≥𝟎 𝒈 −𝟏 𝒙 𝒇 −𝟏 𝒙

6-1 Square Functions as Inverses 6. Justify Mathematical Arguments (1)(G) A classmate says that 𝑓∘𝑔 −1 𝑥 = 𝑓 −1 ∘ 𝑔 −1 𝑥 . Show that this is incorrect by finding examples of f (x) and g (x) for which the equation does not hold. Use at least one square root function in your counterexample. Solution: We need to show that 𝒇∘𝒈 −𝟏 𝒙 ≠ 𝒇 −𝟏 ∘ 𝒈 −𝟏 𝒙 . Let 𝒇 𝒙 = 𝒙 , 𝒕𝒉𝒆𝒏 𝒇 −𝟏 = 𝒙 𝟐 ;𝐚𝐧𝐝 𝒈 𝒙 =𝒙+𝟏 𝒕𝒉𝒆𝒏 𝒈 −𝟏 𝒙 =𝒙−𝟏. 𝐒𝐨𝐥𝐯𝐞 𝐟𝐨𝐫 𝒇∘𝒈 −𝟏 𝒙 : 𝐒𝐨𝐥𝐯𝐞 𝐟𝐨𝐫 𝒇 −𝟏 ∘ 𝒈 −𝟏 𝒙 : 𝐁𝐲 𝐜𝐨𝐦𝐩𝐨𝐬𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬, 𝐁𝐲 𝐜𝐨𝐦𝐩𝐨𝐬𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬, 𝒇 −𝟏 ∘ 𝒈 −𝟏 𝒙 = 𝒇 −𝟏 𝒈 −𝟏 𝒙 𝒇∘𝒈 𝒙 = 𝒙+𝟏 ; and = 𝒇 −𝟏 𝒙−𝟏 𝒇∘𝒈 −𝟏 𝒙 = 𝒙 𝟐 −𝟏 = 𝒙−𝟏 𝟐 = 𝒙 𝟐 −𝟐𝒙+𝟏 𝐒𝐢𝐧𝐜𝐞 𝒙 𝟐 −𝟏 ≠ 𝒙 𝟐 −𝟐𝒙+𝟏, ∴ 𝒇∘𝒈 −𝟏 𝒙 ≠ 𝒇 −𝟏 ∘ 𝒈 −𝟏 𝒙 .