Functions
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This chapter discusses 4 concepts: (1) function relations involving domains and ranges, (2) absolute value functions, (3) composite functions using substitution, and (4) inverse functions by making "x" equal to "y" and solving for y. Examples are provided for each concept to demonstrate finding outputs and inverses of simple functions.
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Functions
- 2. This chapter is simple with 4 understanding into it: a) Function relation b) Absolute function c) Composite function d) Inverse function
- 3. a) Function relation Set A is always > Object > Domain Example: Value of x depend on question Square root of 9 = 3 Square root of 49 = 7 Square root of x = 9 So x is 81 9 3 49 7 x 9 The image that connect Range to object only called The images only. range. {3,7,9} Set A Set B Object of 3 = 9 Image of 9 = 3 Set B is always > Image Object of 7 = 49 Image of 49 = 7 > Codomain Object of 9 = x Image of x = 9
- 4. b) Absolute function Example: The absolute value function f(x)= |7-x| is defined for the domain 0< x < 15 Find f(2) This means you should substitute the Modulus “| |” should be value of 2 into the equation. removed. So your x=5 answers should end with (+) sign. Image of 15 |7-x|= 15 x= -8 Range of f(x) So, x=8 corresponding to given Find f(9) domain. This means same substitution. You So it will be will get x= -2. But since it is in 0 < f(x) < 8 modulus form. Your answer must be 2
- 5. c) Composite function Find f(x) Example: The “f” is located inside g:x. So you must use the method of SUBSTITUTION Given g:x is 3x-2 and gf:x is 3x²+4. Find g(2) Since gf:x is 3x²+4, so you must use g:x for 3x-2 Find g²x substitution. You must make the both 3(2)-2 It means gg(x). equation equal. The “x” in g:x should be =4 So, you must substitute changed into f(x). In this way, g:x into x of g:x. gf(x) = 3f(x)-2 3f(x)-2= 3x²+4 Find g(x)=4 3(3x-2)-2 3f(x)=3x²+4+2 3x-2=4 9x-6-2 3f(x)=3x²+6 3x=4+2 9x-8 f(x)=3x²+6 3x=6 g²x=9x-8 3 x=2 f(x)=x²+2
- 6. d) Inverse function Example: So, y 4= x Given h:x is x 4 and kh:x is 2x-19 3 3 Then, y= x+4 3 Next, y=3x+12 Find hˉ¹(x) This means you should inverse the h:x. hˉ¹(x) is 3x+12 When inverse always, > make “x” as “y” > write the normal equation > after the equation is =x Since kh:x is 2x-19, then you must use h:x for inverse.You must do the usual inverse to h:x same as hˉ¹(x). The inverse=3x+12 will be substituted into x of kh:x.You Find k(x) will get k(x) by expanding the equation like below: The “k” is located outside kh:x. So, you must use the method of khˉ¹(x)= 2(3x+12)-19 INVERSE k(x)=6x+24-19 k(x)=6x+5
- 7. d) Inverse function Example: So, y 4= x Given h:x is x 4 and kh:x is 2x-19 3 3 Then, y= x+4 3 Next, y=3x+12 Find hˉ¹(x) This means you should inverse the h:x. hˉ¹(x) is 3x+12 When inverse always, > make “x” as “y” > write the normal equation > after the equation is =x Since kh:x is 2x-19, then you must use h:x for inverse.You must do the usual inverse to h:x same as hˉ¹(x). The inverse=3x+12 will be substituted into x of kh:x.You Find k(x) will get k(x) by expanding the equation like below: The “k” is located outside kh:x. So, you must use the method of khˉ¹(x)= 2(3x+12)-19 INVERSE k(x)=6x+24-19 k(x)=6x+5