## Calculus: Early Transcendentals (2nd Edition)

The solution is $$y=\bigg\{_{-\frac{1}{3}x+3,\quad x\geq3}^{x+1,\quad x<3;}$$
This is an example of a part by part linear function. We will use the fact that the equation of the linear function passing through points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $$y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1$$This is an example of a part by part linear function. We will use the fact that the equation of the linear function passing through points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $$y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1$$ For $x<3$ we can take $(0,1)$ and $(2,3)$ so we get $$y=\frac{3-1}{2-0}(x-0)+1=\frac{2}{2}x+1$$ so we get $$y=x+1.$$ For $x\geq3$ we can take $(3,2)$ and $(6,1)$ to get $$y=\frac{1-2}{6-3}(x-3)+2=\frac{-1}{3}(x-3)+2.$$ This can be rewritten as $$y=-\frac{1}{3}x+3.$$ Putting this together we get $$y=\bigg\{_{-\frac{1}{3}x+3,\quad x\geq3}^{x+1,\quad x<3;}$$