Differential and Integral Calculus| Feliciano and Uy| Exercise 1.1| Problem 10


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]


Solution:

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

\displaystyle =\frac{4}{x^2}

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

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

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

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

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

\displaystyle =\frac{-3x^2-18x-11}{\left(x+3\right)^2}


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