What is the normal of Y=(3-X^2)(X^3-X+1),(2,1) ?

The normal of Y=(3-X^2)(X^3-X+1),(2,1) is

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normal of Y=(3-X^2)(X^3-X+1),(2,1)

normal of

Y = (3 - X^{2}) (X^{3} - X + 1), (2, 1)

X = - 0.67865 \dots Y + 2.35730 \dots or X = 0.07761 \dots Y + 0.84477 \dots or X = 0.08961 \dots Y + 0.82076 \dots or X = 0.45489 \dots Y + 0.09021 \dots or X = - 2.10252 \dots Y + 5.20505 \dots or X = - 0.15717 \dots Y + 1.31434 \dots or X = - 1.09786 \dots Y + 3.19573 \dots

Solution steps

Slope of the normal line at

(2, 1) : m = \frac{- X ^{5} + 4 X ^{3} - X ^{2} - 3 X + 2}{X - 2}

Find the slope of

Y = (3 - X^{2}) (X^{3} - X + 1) : m_{y} = - 5 X^{4} + 12 X^{2} - 2 X - 3

Compute the slope of the normal line:

m = - \frac{1}{- 5 X ^{4} + 12 X ^{2} - 2 X - 3}

Solve

\frac{- X ^{5} + 4 X ^{3} - X ^{2} - 3 X + 2}{X - 2} = - \frac{1}{- 5 X ^{4} + 12 X ^{2} - 2 X - 3} : X \approx 0.86915 \dots, X \approx - 1.79826 \dots, X \approx 1.65885 \dots, X \approx - 1.60832 \dots, X \approx 1.29476 \dots, X \approx - 1.08306 \dots, X \approx - 0.51928 \dots

Find the normal line at

X = 0.86915 \dots : y = - 0.67865 \dots Y + 2.35730 \dots

Find the normal line at

X = - 1.79826 \dots : y = 0.07761 \dots Y + 0.84477 \dots

Find the normal line at

X = 1.65885 \dots : y = 0.08961 \dots Y + 0.82076 \dots

Find the normal line at

X = - 1.60832 \dots : y = 0.45489 \dots Y + 0.09021 \dots

Find the normal line at

X = 1.29476 \dots : y = - 2.10252 \dots Y + 5.20505 \dots

Find the normal line at

X = - 1.08306 \dots : y = - 0.15717 \dots Y + 1.31434 \dots

Find the normal line at

X = - 0.51928 \dots : y = - 1.09786 \dots Y + 3.19573 \dots

X = - 0.67865 \dots Y + 2.35730 \dots or X = 0.07761 \dots Y + 0.84477 \dots or X = 0.08961 \dots Y + 0.82076 \dots or X = 0.45489 \dots Y + 0.09021 \dots or X = - 2.10252 \dots Y + 5.20505 \dots or X = - 0.15717 \dots Y + 1.31434 \dots or X = - 1.09786 \dots Y + 3.19573 \dots

x \to + π lim (\frac{x}{4} - \frac{π}{2})

\frac{d y}{d x} = 2 x y^{2} + 32 x

\frac{d}{d x} (\frac{3}{2} x^{\frac{2}{3}})

\frac{d y}{d x} = \frac{2 x + 1}{2 y - 2}

x \to 0 lim (\frac{x}{5 x ^{3} - x ^{4}})

What is the normal of Y=(3-X^2)(X^3-X+1),(2,1) ?
The normal of Y=(3-X^2)(X^3-X+1),(2,1) is