## Precalculus (6th Edition)

(a) $f(x) = -\frac{x}{3}+4$ (b) $f(3) = 3$
(a) function notation. Solve for $y$: \begin{array}{ccc} &x+3y&=&12 \\&x+3y-x &= &12-x \\&3y&=&-x+12 \\&\frac{3y}{3} &= &\frac{-x+12}{3} \\&y&=&-\frac{x}{3} + \frac{12}{3} \\&y&=&-\frac{x}{3} + 4 \end{array} Let $y=f(x)$. The equation becomes: $$f(x)=-\frac{x}{3} + 4$$ (b) To find $f(3)$, substitute $3$ to $x$ in $f(x)$ to obtain: $f(x)= -\frac{x}{3}+4 \\f(3) = -\frac{3}{3}+4 \\f(3)=-1+4 \\f(3)=3$