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Problem 41 Hard Difficulty
Find $ f \circ g \circ h $.
$ f(x) = \sqrt{x - 3} $ , $ g(x) = x^2 $ , $ h(x) = x^3 + 2 $
Answer
$$\begin{aligned}
(f \circ g \circ h)(x) &=f(g(h(x)))=f\left(g\left(x^{3}+2\right)\right)=f\left[\left(x^{3}+2\right)^{2}\right] \\
&=f\left(x^{6}+4 x^{3}+4\right)=\sqrt{\left(x^{6}+4 x^{3}+4\right)-3}=\sqrt{x^{6}+4 x^{3}+1}
\end{aligned}$$
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Top Calculus 3 Educators
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Video Transcript
here we have functions F, g and H, and we're going to find the composition f of G of h. Another way to write that is with parentheses. F of G of age like this. So we can see from that that h of X is our innermost function and it's going to be substituted into G. And that means we're going to take a X cubed plus two and substituted in for X in the G function. And that gives us G of h of X equals X cubed plus two quantity squared. So that is G of h of X, and that whole thing gets substituted into the F function. So that whole thing goes in for X in the F function. And that's going to result in the square root of X cubed plus two quantity squared minus three. Now perhaps we want to simplify that. And if we do, we're going to end up squaring this X cubed plus two. We're going to multiply it out using the foil method, and that would give us X to the six power plus four X to the third power plus four and then we still have the minus three. So the last thing we want to do is combine the like terms and we have at the square root of X to the sixth power plus four x to the third power plus one.
Oregon State University
Top Calculus 3 Educators
Lily A.
Johns Hopkins University
Kayleah T.
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Michael J.
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Joseph L.
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