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

The tangent of f(x)=-2x^3,(-1,2) is y=-6x-4

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

tangent of

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

y = - 6 x - 4

Solution steps

Find the slope of

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

Find the line with slope m=

- 6

and passing through

(- 1, 2) : y = - 6 x - 4

y = - 6 x - 4

\frac{9}{( s + 3 ) ^{3}}

\frac{\partial}{\partial y} (3 x^{2} y^{2})

y^{^{'}} + \frac{1}{x} y = 3 x^{2}

x \to 3 lim (5 x - 1)

\frac{d y}{d x} = \frac{e ^{x}}{2 + e ^{x}}

What is the tangent of f(x)=-2x^3,(-1,2) ?
The tangent of f(x)=-2x^3,(-1,2) is y=-6x-4