## University Calculus: Early Transcendentals (3rd Edition)

At 2 points $(-1,-5/6)$ and $(4,-5/3)$, $g(x)$ is parallel with the given line.
$$g(x)=\frac{1}{3}x^3-\frac{3}{2}x^2+1$$ 1) Find $g'(x)$: $$g'(x)=(\frac{1}{3}x^3-\frac{3}{2}x^2+1)'=x^2-3x$$ 2) The slope of the tangent line to $g(x)$ at $x=a$ is $g'(a)=a^2-3a$. Now we need to rewrite the equation of the mentioned line $$8x-2y=1$$ $$2y=8x-1$$ $$y=4x-\frac{1}{2}$$ We know that parallel lines have the same slope. Therefore, all the tangent lines to $g(x)$ parallel to $y=4x-1/2$ must have $g'(a)=4$. In other words, in order to find the required points, we now have to find $a$ such that $g'(a)=a^2-3a=4$: $$a^2-3a=4$$ $$a^2-3a-4=0$$ $$(a+1)(a-4)=0$$ $$a=-1\hspace{1cm}\text{or}\hspace{1cm}a=4$$ - For $a=-1$: $$g(a)=\frac{1}{3}\times(-1)^3-\frac{3}{2}\times(-1)^2+1=-\frac{1}{3}-\frac{3}{2}+1=-\frac{5}{6}$$ - For $a=4$: $$g(a)=\frac{1}{3}\times(4)^3-\frac{3}{2}\times(4)^2+1=\frac{64}{3}-24+1=-\frac{5}{3}$$ Therefore, at 2 points $(-1,-5/6)$ and $(4,-5/3)$, $g(x)$ is parallel with the given line.