Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises - Page 152: 54

\[F'\left( 2 \right)=-2\]

\[\begin{align} & \text{Defining the functions of the graph} \\ & \\ & f\left( x \right)\text{ for }x<3\text{ is given by:} \\ & y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ & \text{Using the points }\left( 1,7 \right)\text{ and }\left( 3,1 \right) \\ & m=\frac{1-7}{3-1}=-3 \\ & y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ & y-1=-3\left( x-3 \right) \\ & y-1=-3x+9 \\ & y=-3x+10 \\ & f\left( x \right)\text{ for }x>3\text{ is given by:} \\ & y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ & \text{Using the points }\left( 3,1 \right)\text{ and }\left( 7,5 \right) \\ & m=\frac{5-1}{7-3}=1 \\ & y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ & y-1=\left( x-3 \right) \\ & y-1=x-3 \\ & y=x-2 \\ & \\ & g\left( x \right)\text{ for }x<3\text{ is given by:} \\ & y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ & \text{Using the points }\left( 3,4 \right)\text{ and }\left( 0,1 \right) \\ & m=\frac{1-4}{0-3}=1 \\ & y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ & y-1=1\left( x-0 \right) \\ & y-1=x+1 \\ & g\left( x \right)\text{ for }x>3\text{ is given by:} \\ & y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ & \text{Using the points }\left( 3,4 \right)\text{ and }\left( 7,0 \right) \\ & m=\frac{0-4}{7-3}=-1 \\ & y-{{y}_{1}}=m\left( x-{{x}_{1}} \right) \\ & y-0=-\left( x-7 \right) \\ & y=-x+7 \\ & \\ & \text{We know that }F=f+g \\ & F'\left( x \right)=\frac{d}{dx}\left[ f\left( x \right)+g\left( x \right) \right] \\ & \text{Using the functions for }x<3 \\ & F'\left( x \right)=\frac{d}{dx}\left[ \left( -3x+10 \right)+\left( x+1 \right) \right] \\ & F'\left( x \right)=\frac{d}{dx}\left[ -2x+11 \right] \\ & F'\left( x \right)=-2 \\ & \text{Calculate }F'\left( 2 \right) \\ & F'\left( 2 \right)=-2 \\ \end{align}\]