Question

How do you write two functions for which `(f*g)(x)=2x^2+11x-6`?

Answer 1

See below.

Explanation:

The trivial response is to define `g(x) = x` and `f(x)=2x^2+11x-6` then

`(f@g)(x)=2x^2+11x-6`

You can also proceed in a more general way making

`g(x) = a x + b` and
`f(x) = x^2+x + c` and then

`(f@g)(x)=a^2x^2+a(1+2b)x+b+b^2+c` and then making an equivalence of coefficients

`{(2 - a^2=0),(11 - a(1 + 2 b)=0),(6 + b + b^2 + c=0):}`

and solving for `a,b,c` giving

`a = sqrt[2], b = 1/4 (11 sqrt[2]-2), c = -167/8` or

`g(x)= sqrt[2] x+1/4 (11 sqrt[2]-2)`
`f(x)=x^2+x-167/8`

Answer 2

`f(x)=(x^2-169)/8` and `g(x)=4x+11`

Explanation:

Another interesting way we could think about this is to complete the square.

`2x^2+11x-6=2(x^2+11/2x)-6`

`color(white)(2x^2+11x-6)=2(x^2+11/2x+121/16)-6-121/8`

`color(white)(2x^2+11x-6)=2(x+11/4)^2-169/8`

So, we can see that our "inner function" is `x+11/4` and the outer function would be `2x^2-169/8`.

So if `f(x)=2x^2-169/8` and `g(x)=x+11/4` then `(f@g)(x)=2x^2+11x-6`.

We can come up with another by just simplifying the function we already had.

`2(x+11/4)^2-169/8=2((4x+11)/4)^2-169/8`

`color(white)(2(x+11/4)^2-169/8)=1/8(4x+11)^2-169/8`

So we can write that for `f(x)=1/8x^2-169/8=(x^2-169)/8` and `g(x)=4x+11`, then `(f@g)(x)=2x^2+11x-6`.

Answer 3

`f(x)=2x^2+3x-20` and `g(x)=x+2`

Explanation:

Another method would be to factor first.

`2x^2+11x-6=2x^2+12x-x-6`

`color(white)(2x^2+11x-6)=(2x-1)(x+6)`

Choose some arbitrary factor, for example, `x+2`. We can express each individual factor in terms of `x+2`.

`color(white)(2x^2+11x-6)=(2(x+2)-5)((x+2)+4)`

If we let `x+2=u` this becomes

`color(white)(2x^2+11x-6)=(2u-5)(u+4)`

`color(white)(2x^2+11x-6)=2u^2+3u-20`

So we can say that if `f(x)=2x^2+3x-20` and `g(x)=x+2`, then `(f@g)(x)=2x^2+11x-6`.