Question

Find $f\circ gandg\circ f$and, where $f\left(x\right)=x^2+1andg\left(x\right)=x+2$ , are functions from $\mathbf{ℝ}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathbf{ℝ}$.

Short answer

The function$f\circ g$ is$(f\circ g)\left(x\right)=x^2+4x+5$

$g\circ f$is$(g\circ f)\left(x\right)=x^2+3$

Step-by-step solution

Step 1:

Composition of f and g :$(f\circ g)\left(a\right)=f\left(g\right(a\left)\right)$

Step 2:

Given:

$g:\mathrm{ℝ}\to \mathrm{ℝ}andf:\mathrm{ℝ}\to \mathrm{ℝ}$

$\begin{array}{r}f\left(x\right)=x^2+1\\ f\left(x\right)=x+2\end{array}$

Since f and g are both from $\mathrm{ℝ}\to \mathrm{ℝ}$,$f\circ gandg\circ f$ are also functions from$\mathrm{ℝ}\to \mathrm{ℝ}$

Use the definition of composition:

$\begin{array}{r}(f\circ g)\left(x\right)=f\left(g\right(x\left)\right)=f(x+2)=(x+2{)}^2+1=x^2+4x+5\\ (g\circ f)\left(x\right)=g\left(f\right(x\left)\right)=g\left(x^2+1\right)=\left(x^2+1\right)+2=x^2+3\end{array}$

Hence, the solution is $\begin{array}{c}(f\circ g)\left(x\right)=x^2+4x+5\\ (g\circ f)\left(x\right)=x^2+3\end{array}$