since we have (5,3), f(5)=3
f'(5) is the slope of the line at x=5
we are given that the line tangent to f(x) at x=5 passes through the point (0,2)
so find the slope of that line
we know it passes through (5,3) and (0,2)
slope between [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is
[tex]\frac{y_1-y_2}{x_1-x_2}[/tex]
slope between (5,3) and (0,2) is [/tex]\frac{3-2}{5-0}=\frac{1}{5}[/tex]
f(5)=3
f'(5)=[tex]\frac{1}{5}[/tex]
