What is the normal of f(x)=-3x^3+4,\at x=-1 ?

The normal of f(x)=-3x^3+4,\at x=-1 is f(x)= 1/9 x+64/9

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normal of f(x)=-3x^3+4,\at x=-1

normal of

f (x) = - 3 x^{3} + 4, at x = - 1

f (x) = \frac{1}{9} x + \frac{64}{9}

Solution steps

Find the tangent point:

(- 1, 7)

Find the slope of

f (x) = - 3 x^{3} + 4 : \frac{df ( x )}{d x} = - 9 x^{2}

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, 7) : f (x) = \frac{1}{9} x + \frac{64}{9}

f (x) = \frac{1}{9} x + \frac{64}{9}

\frac{d}{d x} (nπ x)

\frac{d}{d x} (\frac{2}{1 + x})

\int (- 9.8) d x

x \to - 4 - lim (\frac{x}{( x ^{2} - 16 )})

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

What is the normal of f(x)=-3x^3+4,\at x=-1 ?
The normal of f(x)=-3x^3+4,\at x=-1 is f(x)= 1/9 x+64/9