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

a) There are 2 horizontal tangents. - The first one is $y=-4$. Its perpendicular line at the point of tangency is $x=1$. - The second one is $y=0$. Its perpendicular line at the point of tangency is $x=-1$. b) The smallest slope on the curve is $-3$, acquired at $(0,-2)$. Its perpendicular line at $(0,-2)$ is $y=\frac{1}{3}x-2$.
$$y=f(x)=x^3-3x-2$$ We have the derivative of $f(x)$: $$f'(x)=3x^2-3$$ a) Horizontal tangents are those parallel with the $x$-axis, meaning that their slopes equal $0$. So to find these horizontal tangents, we need to find $x$ so that $f'(x)=0$. $$3x^2-3=0$$ $$x^2-1=0$$ $$x^2=1$$ $$x=\pm1$$ - For $x=1$, $f(1)=1^3-3\times1-2=1-3-2=-4$ So the point of tangency is $(1,-4)$. The equation of the horizontal tangent at $(1,-4)$ is $$y-(-4)=0(x-1)=0$$ $$y=-4$$ A line perpendicular with the horizontal line $y=-4$ at $(1,-4)$ can only the vertical line $x=1$. - For $x=-1$, $f(-1)=(-1)^3-3\times(-1)-2=-1+3-2=0$ So the point of tangency is $(-1,0)$. The equation of the horizontal tangent at $(-1,0)$ is $$y-0=0(x+1)=0$$ $$y=0$$ Again, the only line perpendicular with the horizontal line $y=0$ at $(-1,0)$ can only be $x=-1$. b) As $x^2\ge0$ for all $x$, $(3x^2-3)\ge(3\times0^2-3)=-3$ for all $x$. Therefore, $\min f'(x)=\min(3x^2-3)=-3$ for $x=0$ And for $x=0$, $f(0)=0^3-3\times0-2=-2$ So the smallest slope on the curve is $-3$, acquired at $(0,-2)$. Let's call the perpendicular line $(p)$ whose slope is $s_p$. Since the product of the slopes of 2 perpendicular lines equals $-1$, we have $$s_p=\frac{-1}{-3}=\frac{1}{3}$$ The equation of $(p)$, which passes the point $(0,-2)$, is $$y+2=\frac{1}{3}(x-0)$$ $$y+2=\frac{1}{3}x$$ $$y=\frac{1}{3}x-2$$