## Calculus: Early Transcendentals 8th Edition

Let the initial distance from the lamp be $2d$. $f(2d)=k(2d)^{-2}=\frac{k}{4d^{2}}$ Final distance = d. $f(d)=kd^{-2}=4\times\frac{k}{4d^{2}}$ $\therefore$ The light is 4 times brighter.