Answer
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 $y=\frac{1}{3}x-2$.
Work Step by Step
$$y=f(x)=x^3-3x-2$$
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 $$y+2=\frac{1}{3}(x-0)$$ $$y+2=\frac{1}{3}x$$ $$y=\frac{1}{3}x-2$$
