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Sum, Difference, Product, Quotient, Composition and Inverse

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Presentation on theme: "Sum, Difference, Product, Quotient, Composition and Inverse"β€” Presentation transcript:

1 Sum, Difference, Product, Quotient, Composition and Inverse
Combining Functions Sum, Difference, Product, Quotient, Composition and Inverse

2 Warm Up: 10/7 𝑓 π‘₯ =3 π‘₯ 2 +π‘₯βˆ’5 g π‘₯ = βˆ’2π‘₯+6 βˆ’4 Find 𝑓(βˆ’2).
𝑓 π‘₯ =3 βˆ’ βˆ’2 βˆ’5=5 g π‘₯ = βˆ’2π‘₯+6 βˆ’4 Find g(1). g π‘₯ = βˆ’ βˆ’1=2βˆ’4=βˆ’2 When the transformation is on the inside it affects the... When the transformation is on the outside it affects the…

3 Sum 𝑓+𝑔 π‘₯ =𝑓 π‘₯ +𝑔(π‘₯) Example: 𝑓 π‘₯ =2π‘₯βˆ’1 𝑔 π‘₯ = π‘₯ 2 2π‘₯βˆ’1 + π‘₯ 2
𝑓+𝑔 π‘₯ =𝑓 π‘₯ +𝑔(π‘₯) Example: 𝑓 π‘₯ =2π‘₯βˆ’ 𝑔 π‘₯ = π‘₯ 2 2π‘₯βˆ’1 + π‘₯ 2 𝑓+𝑔 π‘₯ = π‘₯ 2 +2π‘₯βˆ’1

4 Difference π‘“βˆ’π‘” π‘₯ =𝑓 π‘₯ βˆ’π‘”(π‘₯) Example: 𝑓 π‘₯ =2π‘₯βˆ’1 𝑔 π‘₯ = π‘₯ 2 2π‘₯βˆ’1 βˆ’ π‘₯ 2
π‘“βˆ’π‘” π‘₯ =𝑓 π‘₯ βˆ’π‘”(π‘₯) Example: 𝑓 π‘₯ =2π‘₯βˆ’ 𝑔 π‘₯ = π‘₯ 2 2π‘₯βˆ’1 βˆ’ π‘₯ 2 π‘“βˆ’π‘” π‘₯ = βˆ’π‘₯ 2 +2π‘₯βˆ’1

5 Product π‘“βˆ—π‘” π‘₯ =𝑓 π‘₯ βˆ—π‘”(π‘₯) Example: 𝑓 π‘₯ =2π‘₯βˆ’1 𝑔 π‘₯ = π‘₯ 2 2π‘₯βˆ’1 βˆ— π‘₯ 2
π‘“βˆ—π‘” π‘₯ =𝑓 π‘₯ βˆ—π‘”(π‘₯) Example: 𝑓 π‘₯ =2π‘₯βˆ’ 𝑔 π‘₯ = π‘₯ 2 2π‘₯βˆ’1 βˆ— π‘₯ 2 π‘“βˆ—π‘” π‘₯ =2 π‘₯ 3 βˆ’ π‘₯ 2

6 *(you can leave it as: πŸπ’™βˆ’πŸ 𝒙 𝟐 )
Quotient 𝑓 𝑔 π‘₯ = 𝑓 π‘₯ 𝑔 π‘₯ , provided that 𝑔(π‘₯)β‰ 0. Example: 𝑓 π‘₯ =2π‘₯βˆ’ 𝑔 π‘₯ = π‘₯ 2 2π‘₯βˆ’1 π‘₯ 2 = 2π‘₯ π‘₯ 2 βˆ’ 1 π‘₯ 2 𝑓 𝑔 π‘₯ = 2 π‘₯ βˆ’ 1 π‘₯ 2 *(you can leave it as: πŸπ’™βˆ’πŸ 𝒙 𝟐 )

7 Composition of Functions
𝑓°𝑔 π‘₯ =𝑓(𝑔 π‘₯ ) Example: 𝑓 π‘₯ =2π‘₯βˆ’ 𝑔 π‘₯ = π‘₯ 2 𝑓°𝑔 π‘₯ =2 π‘₯ 2 βˆ’1 𝑔°𝑓 π‘₯ = (2π‘₯βˆ’1) 2 =4 π‘₯ 2 βˆ’4π‘₯+1

8 Inverse Function 𝑓 βˆ’1 𝑏 =π‘Ž if and only if 𝑓 π‘Ž =𝑏
If 𝑓 is a one-to-one function with domain 𝐷 and range 𝑅, then the inverse function of 𝒇, denoted 𝑓 βˆ’1 , is the function with domain 𝑅 and range 𝐷 defined by: 𝑓 βˆ’1 𝑏 =π‘Ž if and only if 𝑓 π‘Ž =𝑏 One-to-one: A function in which each element of the range corresponds to exactly one element in the domain.

9 Inverses Algebraically Graphically
Find an equation for 𝑓 βˆ’1 π‘₯ if 𝑓 π‘₯ = π‘₯ π‘₯+1 . Step 1: Switch π‘₯ and 𝑦. π‘₯= 𝑦 𝑦+1 Step 2: Solve for 𝑦. π‘₯ 𝑦+1 =𝑦 π‘₯𝑦+π‘₯=𝑦 π‘₯=π‘¦βˆ’π‘₯𝑦 π‘₯=𝑦 1βˆ’π‘₯ 𝑦= π‘₯ 1βˆ’π‘₯ = 𝑓 βˆ’1 π‘₯ The points (π‘Ž,𝑏) and (𝑏,π‘Ž) in the coordinate plane are symmetric with respect to the line π’š=𝒙. The points (π‘Ž,𝑏) and (𝑏,π‘Ž) are reflections of each other in the line π’š=𝒙.

10 Looking at Inverses Graphically

11 Looking at Inverses Graphically

12 Inverses: reflected over the line π’š=𝒙.
𝑦= 𝑒 π‘₯ and y=ln⁑(π‘₯)

13 Thus, 𝑓(π‘₯) and 𝑔(π‘₯) are inverses.
𝒇 𝒙 = 𝒙 πŸ‘ +𝟏 π’ˆ 𝒙 = πŸ‘ π’™βˆ’πŸ 𝑓 𝑔 π‘₯ = ? = 3 π‘₯βˆ’ =π‘₯βˆ’1+1 =π‘₯ g 𝑓 π‘₯ = ? = 3 ( π‘₯ 3 +1)βˆ’1 = 3 π‘₯ 3 =π‘₯ Thus, 𝑓(π‘₯) and 𝑔(π‘₯) are inverses.


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