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

The tangent of f(x)=-3x^3+4,\at x=-1 is y=-9x-2

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

tangent of

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

y = - 9 x - 2

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}

Compute the slope of

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

x = - 1 : m = - 9

Find the line with slope m=

- 9

and passing through

(- 1, 7) : y = - 9 x - 2

y = - 9 x - 2

\frac{\partial}{\partial x} (m x^{b})

\int_{0}^{3} x 10 - x^{2} d x

\frac{8 x}{7 - tan ( x )}

x \to \infty lim (\frac{- 1}{2})

f (x) = x + 1 (1 - 5 x^{2})^{8}

What is the tangent of f(x)=-3x^3+4,\at x=-1 ?
The tangent of f(x)=-3x^3+4,\at x=-1 is y=-9x-2