Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 10

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PROBLEM:

If  \displaystyle f\left(x\right)=\frac{4}{x+3} and \displaystyle \:g\left(x\right)=x^2-3 , find \displaystyle f\left[g\left(x\right)\right] and \displaystyle g\left[f\left(x\right)\right].


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SOLUTION:

Part A

\begin{align*}

f\left[g\left(x\right)\right] & =\frac{4}{\left(x^2-3\right)+3}\\

& =\frac{4}{x^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}

Part B

\begin{align*}

g\left[f\left(x\right)\right] & =\left(\frac{4}{x+3}\right)^2-3\\

& =\frac{16}{\left(x+3\right)^2}-3\\

& =\frac{16-3\left(x+3\right)^2}{\left(x+3\right)^2}\\

& =\frac{16-3\left(x^2+6x+9\right)}{\left(x+3\right)^2}\\

& =\frac{16-3x^2-18x-27}{\left(x+3\right)^2}\\

& =\frac{-3x^2-18x-11}{\left(x+3\right)^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}

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