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Published byLesley Hines Modified over 6 years ago
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Sum, Difference, Product, Quotient, Composition and Inverse
Combining Functions Sum, Difference, Product, Quotient, Composition and Inverse
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Warm Up: 10/7 π π₯ =3 π₯ 2 +π₯β5 g π₯ = β2π₯+6 β4 Find π(β2).
π π₯ =3 β β2 β5=5 g π₯ = β2π₯+6 β4 Find g(1). g π₯ = β β1=2β4=β2 When the transformation is on the inside it affects the... When the transformation is on the outside it affects theβ¦
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Sum π+π π₯ =π π₯ +π(π₯) Example: π π₯ =2π₯β1 π π₯ = π₯ 2 2π₯β1 + π₯ 2
π+π π₯ =π π₯ +π(π₯) Example: π π₯ =2π₯β π π₯ = π₯ 2 2π₯β1 + π₯ 2 π+π π₯ = π₯ 2 +2π₯β1
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Difference πβπ π₯ =π π₯ βπ(π₯) Example: π π₯ =2π₯β1 π π₯ = π₯ 2 2π₯β1 β π₯ 2
πβπ π₯ =π π₯ βπ(π₯) Example: π π₯ =2π₯β π π₯ = π₯ 2 2π₯β1 β π₯ 2 πβπ π₯ = βπ₯ 2 +2π₯β1
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Product πβπ π₯ =π π₯ βπ(π₯) Example: π π₯ =2π₯β1 π π₯ = π₯ 2 2π₯β1 β π₯ 2
πβπ π₯ =π π₯ βπ(π₯) Example: π π₯ =2π₯β π π₯ = π₯ 2 2π₯β1 β π₯ 2 πβπ π₯ =2 π₯ 3 β π₯ 2
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*(you can leave it as: ππβπ π π )
Quotient π π π₯ = π π₯ π π₯ , provided that π(π₯)β 0. Example: π π₯ =2π₯β π π₯ = π₯ 2 2π₯β1 π₯ 2 = 2π₯ π₯ 2 β 1 π₯ 2 π π π₯ = 2 π₯ β 1 π₯ 2 *(you can leave it as: ππβπ π π )
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Composition of Functions
πΒ°π π₯ =π(π π₯ ) Example: π π₯ =2π₯β π π₯ = π₯ 2 πΒ°π π₯ =2 π₯ 2 β1 πΒ°π π₯ = (2π₯β1) 2 =4 π₯ 2 β4π₯+1
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Inverse Function π β1 π =π if and only if π π =π
If π is a one-to-one function with domain π· and range π
, then the inverse function of π, denoted π β1 , is the function with domain π
and range π· defined by: π β1 π =π if and only if π π =π One-to-one: A function in which each element of the range corresponds to exactly one element in the domain.
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Inverses Algebraically Graphically
Find an equation for π β1 π₯ if π π₯ = π₯ π₯+1 . Step 1: Switch π₯ and π¦. π₯= π¦ π¦+1 Step 2: Solve for π¦. π₯ π¦+1 =π¦ π₯π¦+π₯=π¦ π₯=π¦βπ₯π¦ π₯=π¦ 1βπ₯ π¦= π₯ 1βπ₯ = π β1 π₯ The points (π,π) and (π,π) in the coordinate plane are symmetric with respect to the line π=π. The points (π,π) and (π,π) are reflections of each other in the line π=π.
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Looking at Inverses Graphically
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Looking at Inverses Graphically
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Inverses: reflected over the line π=π.
π¦= π π₯ and y=lnβ‘(π₯)
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Thus, π(π₯) and π(π₯) are inverses.
π π = π π +π π π = π πβπ π π π₯ = ? = 3 π₯β =π₯β1+1 =π₯ g π π₯ = ? = 3 ( π₯ 3 +1)β1 = 3 π₯ 3 =π₯ Thus, π(π₯) and π(π₯) are inverses.
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