MHB 1.1.205 AP calculus exam practice question

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To find \( f(g(-2)) \) where \( f(x) = x - 2 \) and \( g(x) = \frac{x}{x^2 + 1} \), first calculate \( g(-2) \), resulting in \( \frac{-2}{5} \). Then, substitute this value into \( f \), leading to \( f\left(\frac{-2}{5}\right) = \frac{-12}{5} \), which corresponds to option (E). The discussion also touches on formatting issues, such as inserting horizontal lines and using hide/show options for solutions. Overall, the solution process is confirmed to be correct, with some minor requests for formatting improvements.
karush
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given that
$\left[
f(x)=x-2,
\quad
g(x)=\dfrac{x}{x^2+1}\right]$
find $f(g(-2))$
(A) $\dfrac{-11}{5}$
(B) $\dfrac{-4}{17}$
(C) $-3$
(D) $\dfrac{14}{85}$
(E) $\dfrac{-12}{5}$
_____________________________________________________________________________

Solution
find $g(-2)$
$$\dfrac{-2}{(2)^2+1}
=\dfrac{-2}{5}$$
then solve $f(-2/5)$
$$\dfrac{-2}{5}-2
=\dfrac{-2}{5}-\dfrac{10}{5}
=\dfrac{-12}{5}\quad (E)$$

hopefully ok ... typos ... suggestions

is it possible to draw a horizonal line here \hline or \hrule not

also how do you use the hide/show option if you want to hide the solution
 
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karush said:
given that
$\left[
f(x)=x-2,
\quad
g(x)=\dfrac{x}{x^2+1}\right]$
find $f(g(-2))$
(A) $\dfrac{-11}{5}$
(B) $\dfrac{-4}{17}$
(C) $-3$
(D) $\dfrac{14}{85}$
(E) $\dfrac{-12}{5}$
_____________________________________________________________________________

Solution
find $g(-2)$
$$\dfrac{-2}{({\color{red}-2})^2+1}
=\dfrac{-2}{5}$$
then solve $f(-2/5)$
$$\dfrac{-2}{5}-2
=\dfrac{-2}{5}-\dfrac{10}{5}
=\dfrac{-12}{5}\quad (E)$$

hopefully ok ... typos ... suggestions

See stuff in red - otherwise looks good. :)
karush said:
also how do you use the hide/show option if you want to hide the solution

Use [sp]...[/sp] tags.
 
I don't see red stuff
 
karush said:
I don't see red stuff
Put your mouse in the quote box and scroll down.

-Dan
 

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