What is the normal of f(x)=-x^3+4x^2-1,\at x=1 ?

The normal of f(x)=-x^3+4x^2-1,\at x=1 is f(x)=-1/5 x+11/5

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normal of f(x)=-x^3+4x^2-1,\at x=1

normal of

f (x) = - x^{3} + 4 x^{2} - 1, at x = 1

f (x) = - \frac{1}{5} x + \frac{11}{5}

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = - x^{3} + 4 x^{2} - 1 : \frac{df ( x )}{d x} = - 3 x^{2} + 8 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 = 5

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{5}

Find the line with slope m=

- \frac{1}{5}

and passing through

(1, 2) : f (x) = - \frac{1}{5} x + \frac{11}{5}

f (x) = - \frac{1}{5} x + \frac{11}{5}

\frac{d}{d x} (e^{- 3 x + 4})

f (x) = x^{23} + 50 x^{17} + 223

\int_{1}^{\infty} x^{- \frac{1}{7}} d x

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

\frac{d}{d x} (0.01 x^{3} - 0.45 x^{2} + 2.43 x + 300)

What is the normal of f(x)=-x^3+4x^2-1,\at x=1 ?
The normal of f(x)=-x^3+4x^2-1,\at x=1 is f(x)=-1/5 x+11/5