[SOLVED]Q1 Can you pass this 3 question AP calculus Quiz in 10 minutes

karush

Well-known member
1. $f(x)=(2x+1)^3$ and let g be the inverse function of f. Given that$f(0)=1$ what is the value of $g'(1)$?

A $-\dfrac{2}{27}$ B $\dfrac{1}{54}$ C $\dfrac{1}{27}$ D $\dfrac{1}{6}$ E 6

$f(x) = (2x+1)^3 \implies f'(x) = 6(2x+1)^2$
$f$ and $g$ are inverses tells us two things ...
(1) $f(0) = 1 \implies g(1)=0$
(2) $f[g(x)] = x$
take the derivative of equation (2) ...
$f'[g(x)] \cdot g'(x) = 1 \implies g'(x) = \dfrac{1}{f'[g(x)]}$
$g'(1)=\dfrac{1}{f'[g(1)]} = \dfrac{1}{f'(0)} = \dfrac{1}{6}$
2. given that $\left[f(x)=x-2,\quad g(x)=\dfrac{x}{x^2+1}\right]$
find $f(g(-2))$
A $\dfrac{-11}{5}\quad$ B $\dfrac{-4}{17}\quad$ C $-3\quad$ D $\dfrac{14}{85}\quad$ E $\dfrac{-12}{5}$
find $g(-2)$
$$\dfrac{-2}{(2)^2+1}=\dfrac{-2}{5}$$
then solve $f(-2/5)$
$$\dfrac{-2}{5}-2 =\dfrac{-2}{5}-\dfrac{-10}{5} =\dfrac{-12}{5}$$
3. The function f is defined by $f(x)=\dfrac{x}{x+2}$ What points $(x,y)$ on the graph of $f$ have the property that the line tangent to $f$ at $(x,y)$ has slope $\dfrac{1}{2}$?
$$\textsf{A (0,0) only B \left(\dfrac{1}{2},\dfrac{1}{5}\right) only C (0,0) and (-4,2) D (0,0)and \left(4,\dfrac{2}{3}\right) E DNE}$$

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