2b8ddf…: “Except a light calculation”

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1 : Consider the following functions: $f(x) = \sage {p1func1}$ and $g(x) = \sage {p1func2}$ . Compute:

$f(g(x)) = \answer {\sage {p1ans1}}$

$g(f(x)) = \answer {\sage {p1ans2}}$

$(f\circ g)(x) = \answer {\sage {p1ans1}}$

$((g\circ f)(x) = \answer {\sage {p1ans2}}$

You only need to compose the functions you do not need to simplify them. For example, if $f(x) = 13x^2 + x$ and $g(x) = \sqrt {x+1}$ then for $f(g(x))$ you would just copy in the “ $g(x)$ ” wherever there is an “ $x$ ” in $f(x)$ and leave it like that; i.e. you would input “ $13(\sqrt {x+1})^2 + \sqrt {x+1}$ . No need to simplify, reduce, or do anything else!

2 : Consider the following functions: $f(x) = \sage {p2func1}$ and $g(x) = \sage {p2func2}$ . Compute:

$f(g(x)) = \answer {\sage {p2ans1}}$

$g(f(x)) = \answer {\sage {p2ans2}}$

$(f\circ g)(x) = \answer {\sage {p2ans1}}$

$((g\circ f)(x) = \answer {\sage {p2ans2}}$

3 : Consider the following functions: $f(x) = \sage {p3func1}$ and $g(x) = \sage {p3func2}$ . Compute:

$f(g(x)) = \answer {\sage {p3ans1}}$

$g(f(x)) = \answer {\sage {p3ans2}}$

$(f\circ g)(x) = \answer {\sage {p3ans1}}$

$((g\circ f)(x) = \answer {\sage {p3ans2}}$

4 : Consider the following functions: $f(x) = \sage {p4func1}$ and $g(x) = \sage {p4func2}$ . Compute:

$f(g(x)) = \answer {\sage {p4ans1}}$

$g(f(x)) = \answer {\sage {p4ans2}}$

$(f\circ g)(x) = \answer {\sage {p4ans1}}$

$((g\circ f)(x) = \answer {\sage {p4ans2}}$