If f is a linear function such that f(-1) = 5 and f(2) = -1,then f(x + 1) = ?

3 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should notice that the function is linear, hence, substituting x+1 for x in equation of a linear function yields:

`f(x+1) = m(x+1)+n =gt f(x+1) = mx + m + n`

Substituting -1 for `x+1` , or `x = -2` , in equation above yields:

`f(-2) = -2m + m + n =gt 5 = -m + n`

Substituting 2 for `x+1` , or `x = 2-1=1` , in equation above yields:

`f(1)=m+m+n = 2m+n =gt 2m+n = -1`

Subtracting the equation `-m+n=5`  from `2m+n = -1`  yields:

2m+n+m-n=-1-5 => 3m = -6 => m = -2

Substituting -2 for m in equation -m+n=5 yields `2+n=5 =gt n=5-2 =gt n=3` Hence, you need to substitute -2 for m and 3 for n in equation `f(x+1) = -2x - 2+ 3 =gt f(x+1) = -2x + 1`

Hence, evaluating the equation of linear function yields `f(x+1) = -2x + 1.`

tonys538's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

The general equation of a linear function is f(x) = ax + b.

It is known that f(-1) = 5 and f(2) = -1.

This gives two equations that can be solved for a and b.

a*-1 + b = 5 and a*2 + b = -1

-a + b = 5...(1)

2a + b = -1 ...(2)

Subtract (2) from (1)

-a - 2a + b - b = 5 + 1

-3a = 6

a = -2

As -a + b = 5

2 + b = 5

b = 3

The function f(x) = -2x + 3

f(-1) = -2*-1 + 3 = 5

f(2) = -2*2 + 3 = -1

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Let f(x)=ax+b be the linear function.

We'll calculate f(-1):

f(-1) = -a+b

But f(-1)=5 => -a+b=5 (1)

We'll calculate f(2):

f(2)=2a+b

But f(2)=-1 => 2a+b=-1 (2)

We'll subtract (1) from (2):

2a+b+a-b=-1-5

We'll combine like terms:

3a=-6=? a=-6/3 => a=-2

2+b=5 => b=3

The linear function is f(x)=-2x+3

We'll determine f(x+1):

f(x+1)=-2(x+1)+3

f(x+1)=-2x-2+3

f(x+1)=-2x+1 The requested expression of the function f(x+1) is also a linear function: f(x+1)=-2x+1.

We’ve answered 318,915 questions. We can answer yours, too.

Ask a question