Find the equation to the line perpendicular to the tangent to the curve $y=x^3−4x+ 7$ at the point $(2,7)$.

Question

Question

1. Find the equation to the line perpendicular to the tangent to the curve $y=x^3−4x+ 7$ at the point $(2,7)$
2. what's the smallest slope on the curve? at what point on the curve does the curve have the slope.
3. Find the equations for the tangents to the curve at points where the slope of curve is 8.

My work

1. the equation is $y=-\frac18x+\frac{29}4$.
2. the smallest slope on the curve is ____?

Answer 1

Point a)

$y'=3x^2-4\implies y'(2)=8 \implies$ perpendicular line to tangent: $(y-7)=-\frac18(x-2)$

Point b)

let minimize $y'=3x^2-4$

Point c)

set $y'=3x^2-4=8$ to find x

Answer 2

Set the derivative equal to zero... Get $3x^2-4=0\implies x=\pm\frac2 { \sqrt 3}$.
Now evaluate at these points: $(\frac2{\sqrt 3})^3-4\cdot\frac 2{\sqrt 3}+7=\frac8{3 \sqrt3}-\frac8{\sqrt3}+7=-\frac{16}{3\sqrt3}+7$ and $-\frac8{3\sqrt3}+\frac8{\sqrt 3}+7=\frac{16}{3\sqrt3}+7$..

$-\frac{16}{3\sqrt3}+7$ is the minimum...