$\begin{array} { | l | l | l | l | l | }
\hline \mathrm { X } & \mathrm { g } ( \mathrm { x } ) & \mathrm { h } ( \mathrm { x } ) & \mathrm { g } ^ { \prime } ( \mathrm { x } ) & \mathrm { h } ^ { \prime } ( \mathrm { x } ) \
\hline - 3 & 0 & 3 & 1 & 0 \
\hline - 2 & 1 & 2 & 2 & - 3 \
\hline - 1 & 3 & 0 & - 1 & - 2 \
\hline 0 & 2 & 3 & - 2 & 3 \
\hline 1 & 0 & - 1 & - 2 & - 2 \
\hline 2 & - 2 & - 2 & - 1 & 0 \
\hline 3 & - 3 & 0 & 0 & 1 \
\hline
\end{array}$
-Use the table above to find $f ^ { \prime } ( - 1 )$ if $f ( x ) = \frac { g ( x ) h ( x ) + 1 } { g ( x ) }$
A) 19/9
B) - 19/9
C) 17/9
D) - 17/9

Question

$\begin{array} { | l | l | l | l | l | } 
\hline \mathrm { X } & \mathrm { g } ( \mathrm { x } ) & \mathrm { h } ( \mathrm { x } ) & \mathrm { g } ^ { \prime } ( \mathrm { x } ) & \mathrm { h } ^ { \prime } ( \mathrm { x } ) \
\hline - 3 & 0 & 3 & 1 & 0 \
\hline - 2 & 1 & 2 & 2 & - 3 \
\hline - 1 & 3 & 0 & - 1 & - 2 \
\hline 0 & 2 & 3 & - 2 & 3 \
\hline 1 & 0 & - 1 & - 2 & - 2 \
\hline 2 & - 2 & - 2 & - 1 & 0 \
\hline 3 & - 3 & 0 & 0 & 1 \
\hline
\end{array}$
-Use the table above to find $f ^ { \prime } ( - 1 )$ if $f ( x ) = \frac { g ( x ) h ( x ) + 1 } { g ( x ) }$
A) 19/9
B) - 19/9
C) 17/9
D) - 17/9

Quizplus · Accepted Answer

The Answer of $\begin{array} { | l | l | l | l | l | } 
\hline \mathrm { X } & \mathrm { g } ( \mathrm { x } ) & \mathrm { h } ( \mathrm { x } ) & \mathrm { g } ^ { \prime } ( \mathrm { x } ) & \mathrm { h } ^ { \prime } ( \mathrm { x } ) \
\hline - 3 & 0 & 3 & 1 & 0 \
\hline - 2 & 1 & 2 & 2 & - 3 \
\hline - 1 & 3 & 0 & - 1 & - 2 \
\hline 0 & 2 & 3 & - 2 & 3 \
\hline 1 & 0 & - 1 & - 2 & - 2 \
\hline 2 & - 2 & - 2 & - 1 & 0 \
\hline 3 & - 3 & 0 & 0 & 1 \
\hline
\end{array}$
-Use the table above to find $f ^ { \prime } ( - 1 )$ if $f ( x ) = \frac { g ( x ) h ( x ) + 1 } { g ( x ) }$
A) 19/9
B) - 19/9
C) 17/9
D) - 17/9

-Use the Table Above to Find f′(−1)f ^ { \prime } ( - 1 )f′(−1)

-Use the Table Above to Find $f ^ { \prime } ( - 1 )$