What is the normal of f(x)=x^3-3x^2+5,\at x=-1 ?

The normal of f(x)=x^3-3x^2+5,\at x=-1 is f(x)=-1/9 x+8/9

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normal of f(x)=x^3-3x^2+5,\at x=-1

normal of

f (x) = x^{3} - 3 x^{2} + 5, at x = - 1

f (x) = - \frac{1}{9} x + \frac{8}{9}

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

f (x) = x^{3} - 3 x^{2} + 5 : \frac{df ( x )}{d x} = 3 x^{2} - 6 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 9

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{9}

Find the line with slope m=

- \frac{1}{9}

and passing through

(- 1, 1) : f (x) = - \frac{1}{9} x + \frac{8}{9}

f (x) = - \frac{1}{9} x + \frac{8}{9}

What is the normal of f(x)=x^3-3x^2+5,\at x=-1 ?
The normal of f(x)=x^3-3x^2+5,\at x=-1 is f(x)=-1/9 x+8/9