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

The tangent of f(x)=-3x^3-2x^2-x+3,\at x=-1 is y=-6x-1

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

tangent of

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

y = - 6 x - 1

Solution steps

Find the tangent point:

(- 1, 5)

Find the slope of

f (x) = - 3 x^{3} - 2 x^{2} - x + 3 : \frac{df ( x )}{d x} = - 9 x^{2} - 4 x - 1

Compute the slope of

f (x) = - 3 x^{3} - 2 x^{2} - x + 3

x = - 1 : m = - 6

Find the line with slope m=

- 6

and passing through

(- 1, 5) : y = - 6 x - 1

y = - 6 x - 1

\int \frac{x}{( 1 - x ^{2} ) ^{2}} d x

\int_{1}^{2} sin (x) + tan (x) d x

\int (x - 1) (6 x - 5) d x

\frac{d}{d x} (sin^{2} (2 x + 5))

What is the tangent of f(x)=-3x^3-2x^2-x+3,\at x=-1 ?
The tangent of f(x)=-3x^3-2x^2-x+3,\at x=-1 is y=-6x-1