# If f(x)=x+1 and (f@g)(x)=x^2+3x+2, which of the following is g(x)?

## A) x+2 B) ${x}^{2} + 3 x$ C)${x}^{2} + 3 x - 1$ D)${x}^{2} + 3 x + 1$ E)${\left(x + 1\right)}^{2} \left(x + 2\right)$

Mar 17, 2018

D)

#### Explanation:

Proposing

$g \left(x\right) = a {x}^{2} + b x + c$

we have

$f \left(g \left(x\right)\right) = a {x}^{2} + b x + c + 1 = {x}^{2} + 3 x + 2$

now comparing coefficients

$\left\{\begin{matrix}c = 1 \\ b = 3 \\ a = 1\end{matrix}\right.$

hence

$g \left(x\right) = {x}^{2} + 3 x + 1$

Mar 18, 2018

$g \left(x\right) = {x}^{2} + 3 x + 1$

#### Explanation:

Since the function $f \left(x\right)$ is one-to-one and onto, it is invertible. In fact, it is easy to see that the inverse function is

${f}^{-} 1 \left(x\right) = x - 1$

Quick check : $\left({f}^{-} 1 \circ f\right) \left(x\right) = {f}^{-} 1 \left(f \left(x\right)\right) = {f}^{-} 1 \left(x + 1\right) = x$

Thus

$g \left(x\right) = \left(\left({f}^{-} 1 \circ f\right) \circ g\right) \left(x\right) = \left({f}^{-} 1 \circ \left(f \circ g\right)\right) \left(x\right)$
$q \quad = {f}^{-} 1 \left(\left(f \circ g\right) \left(x\right)\right) = {f}^{-} 1 \left({x}^{2} + 3 x + 2\right) = \left({x}^{2} + 3 x + 2\right) - 1$

So $g \left(x\right) = {x}^{2} + 3 x + 1$