What is the normal of y=2x^3-3x+1,(1,0) ?

The normal of y=2x^3-3x+1,(1,0) is y=-1/3 x+1/3

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normal of y=2x^3-3x+1,(1,0)

normal of

y = 2 x^{3} - 3 x + 1, (1, 0)

y = - \frac{1}{3} x + \frac{1}{3}

Solution steps

Find the slope of

y = 2 x^{3} - 3 x + 1 : \frac{d y}{d x} = 6 x^{2} - 3

Compute the slope of

y = 2 x^{3} - 3 x + 1

(1, 0) : m = 3

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{3}

Find the line with slope m=

- \frac{1}{3}

and passing through

(1, 0) : y = - \frac{1}{3} x + \frac{1}{3}

y = - \frac{1}{3} x + \frac{1}{3}

\frac{d}{d x} (- x sin (x) + cos (x))

\int \frac{1}{3 + cos ( x )} d x

\int_{- π}^{π} ∣ sin (x) ∣ d x

y^{^{'}} = (3 x - 2) (30 x + 19)

\frac{\partial}{\partial x} (400 \cdot sin (x) \cdot cos (x))

What is the normal of y=2x^3-3x+1,(1,0) ?
The normal of y=2x^3-3x+1,(1,0) is y=-1/3 x+1/3