Given the function ƒ(x ) = 3x + 1, evaluate ƒ(a + 1).
Question
Updated 1/18/2015 10:29:58 PM
Flagged by yeswey [1/18/2015 10:28:56 PM]
s
Original conversation
User: Given the function ƒ(x ) = 3x + 1, evaluate ƒ(a + 1).

User: If ƒ(x ) = 3x + 1, then ƒ(a + h ) - ƒ(a ) =

Question
Updated 1/18/2015 10:29:58 PM
Flagged by yeswey [1/18/2015 10:28:56 PM]
Rating
3
ƒ(x) = 3x + 1,
ƒ(a + 1) = 3(a + 1) + 1 = 3a + 3 + 1 = 3a + 4
Confirmed by Andrew. [1/18/2015 10:34:18 PM]
3
ƒ(x ) = 3x + 1,
ƒ(a + h ) - ƒ(a )
= 3(a + h) + 1 - (3a + 1)
= 3a + 3h + 1 - 3a - 1
= 3h
Confirmed by Andrew. [1/18/2015 10:40:11 PM]

Questions asked by the same visitor
If ƒ(x ) = x 2 + 1 and g(x ) = 3x + 1, find g(20) + ƒ(6).
Question
Updated 1/18/2015 10:38:09 PM
ƒ(x ) = x^2 + 1 and g(x ) = 3x + 1,
g(20) + ƒ(6)
= 3(20) + 1 + (6^2 + 1)
= 61 + 37
= 98
If ƒ(x ) = x 2 + 1 and g(x ) = 3x + 1, find [ƒ(2) - g(1)] 2. User: If ƒ(x ) = x 2 + 1 and g(x ) = 3x + 1, find ƒ(2) + g(3). User: If ƒ(x ) = x 2 + 1, find ƒ(a + 1).
Question
Updated 1/18/2015 10:35:46 PM
ƒ(x ) = x^2 + 1 and g(x ) = 3x + 1,
[ƒ(2) - g(1)]^2
= {(2^2 + 1) - [3(1) + 1]}^2
= (5 - 4)^2
= 1^2
= 1
Confirmed by Andrew. [1/18/2015 10:41:06 PM]
ƒ(x ) = x^2 + 1 and g(x ) = 3x + 1,
ƒ(2) + g(3)
= (2^2 + 1) + [3(3) + 1]
= 5 + 10
= 15
Confirmed by Andrew. [1/18/2015 10:42:08 PM]
ƒ(x ) = x^2 + 1,
ƒ(a + 1) = (a + 1)^2 + 1 = a^2 + 2a + 2
Confirmed by Andrew. [1/18/2015 10:43:17 PM]
If ƒ(x ) = x 2 + 1 and g(x ) = 3x + 1, find [ƒ(4)] 2. User: If g(x ) = 3x + 1, evaluate g(-14). User: If ƒ(x ) = x 2 + 1 and g(x ) = 3x + 1, find ƒ(-1) + 5g(-3).
Question
Updated 1/18/2015 10:44:28 PM
ƒ(x ) = x^2 + 1 and g(x ) = 3x + 1,
[ƒ(4)]^2 = (4^2 + 1)^2 = 17^2 = 289
Confirmed by Andrew. [1/18/2015 10:52:52 PM]
g(x ) = 3x + 1,
g(-14) = 3(-14) + 1 = -42 + 1 = -41
Confirmed by Andrew. [1/18/2015 11:00:17 PM]
ƒ(x ) = x^2 + 1 and g(x ) = 3x + 1,
ƒ(-1) + 5g(-3)
= [(-1)^2 + 1] + 5[3(-3) + 1]
= 2 + 5(-8)
= 2 - 40
= -38
Confirmed by Andrew. [1/18/2015 11:00:48 PM]
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