Determine all linear function if f(f(x)) = 4x + 3

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(f(x) )= 4x+3.

To find f(x).

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

Then f(f(x) = a* (ax+b) +b = a^2x+ab+b

Therefore  a^2x+(ab+b)  must be equal to  4x+3 as given.

So we equate the coefficients of corresponding powers on both sides:

Therefore a^2 = 4. So a = sqrt4 = 2 or -2.

Equate constants:

ab+b = 3.

When a =2,

ab+b = 2b+b = 3b =3. So b =1.

 So f(x) = 2x+1.

When a = -2,

 constant term is ab+b =  -2b+b = 3, -b = 3. S0 b = -3.

Therefore f(x) = -2x-3.

So,

 f(x) = 2x+1 or

f(x) = -2x -3,

 are the possinble 2 linear functions which are solution to f(f(x)) = 4x+3.

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The standard expression for a linear function is:

f(x) = ax + b, where a>0

We'll impose the constraint from enunciation and we'll get:

f(f(x)) = a*f(x) + b

We'll substitute f(x) by ax+b

f(f(x)) = a(ax+b) + b

We'll remove the brackets and we'll have:

f(f(x)) = a^2*x + ab + b (1)

But f(f(x)) = 4x + 3 (2)

We'll put (1) = (2):

a^2*x + ab + b = 4x + 3

For te expressions to be equal the correspondent coefficients have to be equal:

a^2 = 4

ab + b = 3 => b(a+1) = 3 (3)

Because a>0 => a = sqrt4

a = 2

We'll substitute a in (3):

b(a+1) = 3

b(2+1) = 3

3b = 3

We'll divide by 3:

b = 1

The linear function f(x) = 2x + 1.

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