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

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 $y=8x+17$
$$y=f(x)=x^3-4x+1$$ 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 $$(y-1)=8(x+2)$$ $$y-1=8x+16$$ $$y=8x+17$$