#### Answer

a) The equation of the perpendicular line is $y=-\frac{1}{8}x+\frac{5}{4}$
b) The smallest slope on the curve is $-4$ at the point $(0,1)$.
c) There are 2 tangent lines whose slope is $8$. One is $y=8x-15$. The other is $y=8x+17$

#### Work Step by Step

$$y=f(x)=x^3-4x+1$$
a) Find $f'(x)$: $$f'(x)=(x^3-4x+1)'=3x^2-4+0=3x^2-4$$
So, $$f'(2)=3\times2^2-4=3\times4-4=8$$
Let's call the tangent line to $f(x)$ at $(2,1)$ $(t)$ with slope $s_t$ and the perpendicular line at $(2,1)$ $(p)$ with slope $s_p$.
We have $s_t=f'(2)=8$
Since $(t)$ and $(p)$ are perpendicular with each other, the product of their slopes equals $-1$.
That means $$s_p=\frac{-1}{s_t}=-\frac{1}{8}$$
So the equation of $(p)$, which is perpendicular with $(t)$ at $(2,1)$, is $$(y-1)=-\frac{1}{8}(x-2)$$ $$y-1=-\frac{1}{8}x+\frac{1}{4}$$ $$y=-\frac{1}{8}x+\frac{5}{4}$$
b) The slope of the tangent line to $f(x)$ at $x=a$ is $f'(a)=3a^2-4$.
We have $a^2\ge0$ for all $a$.
Therefore, $(3a^2-4)\ge(3\times0-4)=-4$ for all $a$.
That means $\min(3a^2-4)=-4$ for $a=0$.
For $a=0$, $f(0)=0^3-4\times0+1=1$.
So the smallest slope on the curve is $-4$ at the point $(0,1)$.
c) We know that the slope of the tangent line to $f(x)$ at $x=a$ is $f'(a)=3a^2-4$.
So for a slope to equal $8$, $$3a^2-4=8$$ $$a^2=4$$ $$a=\pm2$$
- With $a=2$: $f(2)=2^3-4\times2+1=8-8+1=1$
So the point of tangency is $(2,1)$.
The equation of the tangent line here is $$(y-1)=8(x-2)$$ $$y-1=8x-16$$ $$y=8x-15$$
- With $a=-2$: $f(-2)=(-2)^3-4\times(-2)+1=-8+8+1=1$
So the point of tangency is $(-2,1)$.
The equation of the tangent line here is $$(y-1)=8(x+2)$$ $$y-1=8x+16$$ $$y=8x+17$$