## Calculus: Early Transcendentals 8th Edition

The equation of the tangent line $(l)$ to graph $g$ at the point where $x=3$ is $$(l): y=-2x+18$$
$f(3)=4$ and $f'(3)=-2$ $$g(x)=xf(x)$$ 1) First, we need to find $g'(3)$ $$g'(x)=x'f(x)+xf'(x)$$ $$g'(x)=f(x)+xf'(x)$$ Then, $$g'(3)=f(3)+3\times f'(3)$$ $$g'(3)=4+3\times(-2)$$ $$g'(3)=-2$$ 2) At $x=3$, then $g(3)=3\times f(3)=3\times4=12$ So, the point where $x=3$ that the tangent line $(l)$ touches graph $g$ is point $A(3,12)$. The slope of the tangent line $(l)$ to the graph $g$ at $A(3,12)$ equals $g'(3)$. Therefore, the equation of the tangent line $(l)$ to graph $g$ at $A(3,12)$ is $$(l): y-12=g'(3)(x-3)$$ $$(l): y-12=-2(x-3)$$ $$(l): y-12=-2x+6$$ $$(l): y=-2x+18$$