Given the function ƒ(x ) = 3x + 1, evaluate ƒ(a + 1).

Question

Asked 1/13/2015 8:28:52 PM

Updated 1/18/2015 10:29:58 PM

2 Answers/Comments

Flagged by yeswey [1/18/2015 10:28:56 PM]

s

Question

Asked 1/13/2015 8:28:52 PM

Updated 1/18/2015 10:29:58 PM

2 Answers/Comments

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

ƒ(a + 1) = 3(a + 1) + 1 = 3a + 3 + 1 = 3a + 4

Added 1/18/2015 10:28:49 PM

This answer has been confirmed as correct and helpful.

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

ƒ(a + h ) - ƒ(a )

= 3(a + h) + 1 - (3a + 1)

= 3a + 3h + 1 - 3a - 1

= 3h

Added 1/18/2015 10:29:58 PM

This answer has been confirmed as correct and helpful.

Confirmed by Andrew. [1/18/2015 10:40:11 PM]

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

1 Answer/Comment

ƒ(x ) = x^2 + 1 and g(x ) = 3x + 1,

g(20) + ƒ(6)

= 3(20) + 1 + (6^2 + 1)

= 61 + 37

= 98

g(20) + ƒ(6)

= 3(20) + 1 + (6^2 + 1)

= 61 + 37

= 98

Added 1/18/2015 10:38:05 PM

This answer has been confirmed as correct and helpful.

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

3 Answers/Comments

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

[ƒ(2) - g(1)]^2

= {(2^2 + 1) - [3(1) + 1]}^2

= (5 - 4)^2

= 1^2

= 1

Added 1/18/2015 10:34:01 PM

This answer has been confirmed as correct and helpful.

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

ƒ(2) + g(3)

= (2^2 + 1) + [3(3) + 1]

= 5 + 10

= 15

Added 1/18/2015 10:35:08 PM

This answer has been confirmed as correct and helpful.

Confirmed by Andrew. [1/18/2015 10:42:08 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

3 Answers/Comments

ƒ(x ) = x^2 + 1 and g(x ) = 3x + 1,

[ƒ(4)]^2 = (4^2 + 1)^2 = 17^2 = 289

[ƒ(4)]^2 = (4^2 + 1)^2 = 17^2 = 289

Added 1/18/2015 10:41:03 PM

This answer has been confirmed as correct and helpful.

Confirmed by Andrew. [1/18/2015 10:52:52 PM]

g(x ) = 3x + 1,

g(-14) = 3(-14) + 1 = -42 + 1 = -41

g(-14) = 3(-14) + 1 = -42 + 1 = -41

Added 1/18/2015 10:41:33 PM

This answer has been confirmed as correct and helpful.

Confirmed by Andrew. [1/18/2015 11:00:17 PM]

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