## Calculus 10th Edition

(a) $f(1)=2, f(4)=5$ (b) $3$ (c) $y=x+1$
(a) $f(x)$ is the same as asking for the y-value of a certain value of x, so for $x=1, y=2$ and for $x=4, y=5$ (the points $(1,2)$ and $(4,5)$ are given). (b) since we got $f(1)$ and $f(4)$ from part a, we can simply plug in the values: $f(4)-f(1)=5-2=3$ (c) again, since we already know that $f(1)=2$ and $f(4)=5$, we can just plug the values into the problem: $y=\frac{f(4)-f(1)}{4-1}(x-1)+f(1)$ $y=\frac{5-2}{4-1}(x-1)+2$ $y=\frac{3}{3}(x-1)+2$ $y=x-1+2$ $y=x+1$+2 y=x+3