What is the normal of y=-(2x-2)^{1/3},(-3,2) ?

The normal of y=-(2x-2)^{1/3},(-3,2) is y=6x+20

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normal of y=-(2x-2)^{1/3},(-3,2)

normal of

y = - (2 x - 2)^{\frac{1}{3}}, (- 3, 2)

y = 6 x + 20

Solution steps

Find the slope of

y = - (2 x - 2)^{\frac{1}{3}} : \frac{d y}{d x} = - \frac{2}{3 ( 2 x - 2 ) ^{\frac{2}{3}}}

Compute the slope of

y = - (2 x - 2)^{\frac{1}{3}}

(- 3, 2) : m = - \frac{1}{6 ( - 1 ) ^{\frac{2}{3}}}

Compute the slope of the perpendicular line:

m_{p} = 6

Find the line with slope m=

6

and passing through

(- 3, 2) : y = 6 x + 20

y = 6 x + 20

\frac{1}{2} \frac{d y}{d x} = y + 1 cos (x)

y = e^{2 x^{2}}

\int (x + x e^{x}) d x

x \to 0 lim ((x^{2} + 1)^{\frac{1}{x}})

\int 5 x^{4} e^{9 x^{5}} d x

What is the normal of y=-(2x-2)^{1/3},(-3,2) ?
The normal of y=-(2x-2)^{1/3},(-3,2) is y=6x+20