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Given $f (x) = \frac{x - 3}{5 x + 1}$ $f(x) = (x-3) / (5x+1)$ and $g (x) = \frac{x - 1}{x^{2}}$ $g(x) = (x-1) / (x^2)$ how do you find f(g(x))?

1 Answer

Shwetank Mauria

May 1, 2016

$f (g (x)) = \frac{- 3 x^{2} + x - 1}{x^{2} + 5 x - 5}$ $f(g(x))=(-3x^2+x-1)/(x^2+5x-5$

Explanation:

$f (x) = \frac{x - 3}{5 x + 1}$ $f(x)=(x-3)/(5x+1)$ and $g (x) = \frac{x - 1}{x^{2}}$ $g(x)=(x-1)/x^2$

So $f (g (x)) = f (\frac{x - 1}{x^{2}}) = \frac{\frac{x - 1}{x^{2}} - 3}{5 (\frac{x - 1}{x^{2}}) + 1}$ $f(g(x))=f((x-1)/x^2)=((x-1)/x^2-3)/(5((x-1)/x^2)+1)$

= $\frac{\frac{x - 1 - 3 x^{2}}{x^{2}}}{\frac{5 (x - 1) + x^{2}}{x^{2}}} = \frac{x - 1 - 3 x^{2}}{x^{2}} \times \frac{x^{2}}{(5 (x - 1) + x^{2})}$ $((x-1-3x^2)/x^2)/((5(x-1)+x^2)/x^2)=(x-1-3x^2)/x^2xxx^2/((5(x-1)+x^2)$

= $\frac{x - 1 - 3 x^{2}}{5 x - 5 + x^{2}} = \frac{- 3 x^{2} + x - 1}{x^{2} + 5 x - 5}$ $(x-1-3x^2)/(5x-5+x^2)=(-3x^2+x-1)/(x^2+5x-5$

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