Question

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

Short answer

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 :$\mathbf{(}\mathit{f}\mathbf{\circ }\mathit{g}\mathbf{)}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{g}\mathbf{\right(}\mathit{a}\mathbf{\left)}\mathbf{\right)}$

Step 2:

Given:

$$\begin{gathered} g:\mathrm{ℝ}\to \mathrm{ℝ}\text{and f}:\mathrm{ℝ}\to \mathrm{ℝ} \\ f\left(x\right)=x^2+1 \\ f\left(x\right)=x+2 \end{gathered}$$

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

Use the definition of composition:

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

Hence, the solution is$(f\circ g)\left(x\right)=x^2+4x+5$

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