- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

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

$f$ 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}$

find $f(g(-2))$

A $\dfrac{-11}{5}\quad $ B $\dfrac{-4}{17}\quad$ C $-3\quad $ D $\dfrac{14}{85}\quad$ E $\dfrac{-12}{5}$

$$\dfrac{-2}{(2)^2+1}=\dfrac{-2}{5}$$

then solve $f(-2/5)$

$$\dfrac{-2}{5}-2

=\dfrac{-2}{5}-\dfrac{-10}{5}

=\dfrac{-12}{5}$$

$$\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}$$

$\dfrac{d}{dx}\dfrac{x}{x+2}=\dfrac{1}{2}

\quad x=-4,0 \quad\therefore y=2,0

\quad x=-4,0 \quad\therefore y=2,0

Note solutions came by replies from MHB forum