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$\frac{1}{4x^4-12x^3+9x^2}$

The functions available are

$f(x) = 2x$

$g(x) = x-1$

$h(x) = x^2$

$i(x) = 2x^2-3x$

$j(x) =\sqrt x$

$k(x) = \frac1x$

I can only get as far as $k(i(x))$ and not really sure how to even go from there. Any solutions or help appreciated. :)

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  • $\begingroup$ What are $h(g(x))$, $h(i(x))$, $i(g(x))$, $i(i(x))$? $\endgroup$ – Trevor Gunn Sep 18 '17 at 4:13
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$$\frac1{4x^4-12x^3+9x^2}=k(4x^4-12x^3+9x^2)$$ $$=k(h(2x^2-3x))=k(h(i(x))$$

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Your idea to use $k,i$ is correct, however you need one more function. This would be $h$, now $$k(h(i(x)) = \frac{1}{(h(i(x))}= \frac{1}{i^2(x)}=\frac{1}{(2x^2-3x)^2}=\frac{1}{((2x^2)^2-2(2x^2)(3x)+(3x)^2)} = \frac{1}{4x^4-12x^3+9x^2}$$

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