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I'm trying to find $f \circ g \circ h$ when $f(x) = -1/(10x), g(x) = -5x^3$ and $h(x) = -4x^2+10$.

The way I did it...

$$f(g(h(x)))= f(g(-4x^2+10))$$

$f(-5x^3(-4x^2-10)$ <--- this part confused me, but I came up with...

$-5x^3(-4x^2-10)/(-10x)$

I know that's completely wrong, but what am I doing wrong and how is this supposed to be solved properly?

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I've changed algebra tag to algebra-precalculus, since we don't use algebra tag anymore, see meta for details. –  Martin Sleziak Sep 26 '12 at 11:00
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3 Answers

up vote 4 down vote accepted

You’re missing the basic idea behind composition. Let’s look just at the first step, $$(g\circ h)(x)=g\big(h(x)\big)\;,$$ where $g(x) = -5x^3$ and $h(x) = -4x^2+10$. The notation $g\big(h(x)\big)$ means that you’re going to use the output of $h$ as the input to $g$. You feed $x$ into $h$, and you $-4x^2+10$ as output. Now you feed that output into $g$:

$$g\big(h(x)\big)=g\left(-4x^2+10\right)\;.$$

What does $g$ do to its input? When the input is simply $x$, it spits out $-5x^3$. In other words, it cubes the input and then multiplies it by $-5$. When the input is $-4x^2+1)$, it’s going to do the same thing: cube it, and then multiply by $-5$. Thus,

$$g\big(h(x)\big)=g\left(-4x^2+10\right)=-5\left(-4x^2+10\right)^3\;.\tag{1}$$

When you’re just starting to work with composition of functions it may help to introduce new variables temporarily. Here, for instance, I could let $u=-4x^2+10$. Then

$$g\left(-4x^2+10\right)=g(u)\;,$$

and it may be easier to see that $g(u)=-5u^3=-5\left(-4x^2+10\right)^3$ than to see it all in one go as in $(1)$.

Now we want to feed $g\big(h(x)\big)$ into $f$:

$$f\left(g\big(h(x)\big)\right)=f\left(-5\left(-4x^2+10\right)^3\right)\;.$$

Now $f(x)=\dfrac1{-10x}$: it multiplies its input by $-10$ and then takes the reciprocal of that. Thus,

$$f\left(g\big(h(x)\big)\right)=f\left(-5\left(-4x^2+10\right)^3\right)=\frac1{-10\left(-5\left(-4x^2+10\right)^3\right)}\;,$$

which can be simplified to

$$\frac1{50\left(-4x^2+10\right)^3}\;.$$

You can even pull out a factor of $2^3=8$ if you wish:

$$\frac1{50\left(-4x^2+10\right)^3}=\frac1{400\left(-2x^2+5\right)}\;.$$

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General composition $f\circ g \circ h(x) $ is the same as $f(g(h(x)))$. Now, for example,$$f(x) = x\\ g(x) = x^2 \\ h(x) = x^3 \\\text{then} \\ \begin{align}f(g(h(x)) & = & f(g(x^3))) \\ & = & f((x^3)^2) \\ & = & f(x^6)\\ & = & x^6 \end{align}$$

Notice: you have learnt that if $f(x) = x^2$, then $f(h) = h^2$, $f(x + h) = (x + h)^2$ and so on. Similarly, $g(h(x)) = g(x^3) = (x^3)^2 = x^6 $.

Use the above example for simplifying your question accordingly.

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$g(h(x))=g(-4x^2+10)=-5(-4x^2+10)^3$.

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