Chapter P - P.3 - Linear Models and Rates of Change - Exercises - Page 30: 108

False

Here is a counterexample: $f(x)=x+1$ $a=4$ If we assume the statement $f(ax)=af(x)$ is true, then we run into a contradiction: $f(ax)=af(x)$ $f(4x)=4(x+1)$ $4x+1=4x+4$ $1=4$ Therefore the statement $f(ax)=af(x)$ is false.