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I'm the novice to calculus,and I'm reading "The cartoon guide to calculus"by LARRY GONICK,there're the problems that I don't know how to compute:

1.Here are some composite functions.Identify their inside and outside components ,and write each given function in the form u(v(x)) or u(v(w(x))).

  • a. $h(x) = 2^{\cos x}$

  • b. $h(x) = \sqrt{\ln(x^2-1)}$

  • c. $h(x) = 4e^{3x}+ e^{2x} + 6e^{x} -99$

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    @Arthur thanks for reminding. – 2017-01-31

2 Answers 2

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I am guessing the question asks you to think of, say, the first function as first transforming (mapping) $x$ to $\cos{x}$ and then mapping $\cos{x}$ to $2^{\cos{x}}$. One way to write this is to write $v(x)=\cos{x}$ and $u(x)=2^{x}$. Then $u(v(x))=u(\cos{x})=2^{\cos{x}}$. You can do the other part, keeping in mind that the number of 'steps' through which $x$ is transformed can be three instead of two (or any other number).

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For $a$, consider the maps $f(x)=\cos(x)$ and $g(x)=2^x$, then $g\circ f(x)= g(\cos(x))=2^{\cos(x)}=h(x)$.

For $b$, consider the maps $f(x)=x^2-1$, $g(x)=\ln(x)$ and $k(x)=\sqrt{x}$. Then $k\circ g\circ f(x)=\sqrt{\ln(x^2-1)}$.

For $c$, consider the maps $f(x)=e^x$ and $g(x)=4x^3+x^2+6x-99$. Then $g\circ f(x)=4(e^{x})^3+(e^{x})^2+6e^x-99=4e^{3x}+e^{2x}+6e^x-99$.