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Zenah Abdulaal
Dai Thai
Chem. 400
Classify the following functions as one of the types of functions that we have discussed.
Quadratic Function
Find an equation of the quadratic below:
Find a formula for a cubic function if (−2) = ?(1) = ?(3) = 0 and ?(2) = 8
f(x) = -x^3 + 6x^2 – 13x + 8
Ex 1) Let f(x) = 2x – 3. Find
- a) f(–5)
- b) f(2)a) f(–5) = 2(-5) – 3
= -10 – 3
= -13
- b) f(2) = 2(2) – 3
= 4 – 3
= 1
Ex 2) Use the vertical line test to identify graphs in which y is a function of x.
- a) b) c) d)
- a) Yes – The vertical line test passes.
- b) No – The vertical line test fails.
- c) Yes – The vertical line test passes.
- d) No – The vertical line test fails.
Ex 3) Determine whether the relations are functions:
- b) c)) 2? + ?2 = 16 ? = ?2 + 1 ? = ????
- a) This is a function because it produces a unique output for each input. The domain is all real numbers, and the range is all real numbers greater than or equal to 4.
- b) This is a function because it produces a unique output for each input. The domain is all real numbers, and the range is all real numbers greater than or equal to 1.
- c) This is a function because it produces a unique output for each input. The domain is all real numbers, and the range is all real numbers greater than or equal to -1.
Ex 4) Let and? ( ) = ?+1 ?−2 ? ?( ) = ? + 2
Find the domain of each function.
f(x) = (x+1)/(x-2)
Domain of f(x) = {x | x ≠ 2}
g(x) = x + 2
Domain of g(x) = All real numbers
Ex 5) Let f(x) = 2x – 3. State the domain of the function, and find:
- a) f(–5)
- b) f(2)
- c) f(a +1)
- d) f(x + h)
Domain: x ∈ R
- a) f(–5) = 2(-5) – 3 = -13
- b) f(2) = 2(2) – 3 = 1
- c) f(a + 1) = 2(a + 1) – 3 = 2a + 1 – 3 = 2a – 2
- d) f(x + h) = 2(x + h) – 3 = 2x + 2h – 3
Ex 6) Simplify the difference quotient for ? ?+ℎ( )−?(?)
ℎ ? ( ) = 3 − ?2
Piecewise Functions:
? ?+ℎ( )−?(?) = (3 − (?+ℎ)2) − (3 − ?2)
= (3 − ?2 − 2?ℎ − ℎ2) − (3 − ?2)
= −2?ℎ − ℎ2
Simplified Difference Quotient:
? ?+ℎ( )−?(?) = −2?ℎ − ℎ2
Ex 7) Graph the piecewise function ? ( ){?2 − 2? + 1 ??? ? < 2 ? − 1| | ??? ? ≥2
For x < 2
f(x) = x2 – 2x + 1
f(0) = 0 – 0 + 1 = 1
f(1) = 1 – 2 + 1 = 0
f(2) = 4 – 4 + 1 = 1
For x ≥ 2
f(x) = x – 1
f(2) = 2 – 1 = 1
f(3) = 3 – 1 = 2
f(4) = 4 – 1 = 3
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