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

The tangent of f(x)=-3x^3-3x^2-x+3,\at x=-1 is y=-4x

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

tangent of

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

y = - 4 x

Solution steps

Find the tangent point:

(- 1, 4)

Find the slope of

f (x) = - 3 x^{3} - 3 x^{2} - x + 3 : \frac{df ( x )}{d x} = - 9 x^{2} - 6 x - 1

Compute the slope of

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

x = - 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(- 1, 4) : y = - 4 x

y = - 4 x

\int e^{3 x - 1} d x

x \to - 4 lim (x^{2} - 5 x + 12)

\int - \frac{sin ( x )}{cos ( x )} d x

\int \frac{3}{e ^{2 x}} d x

\int_{0}^{1} 8 x^{3} d x

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