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

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

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

tangent of

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

y = 3 x + 2

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x^{3}

(- 1, - 1) : m = 3

Find the line with slope m=

3

and passing through

(- 1, - 1) : y = 3 x + 2

y = 3 x + 2

\frac{d y}{d x} = y x

\frac{\partial}{\partial x} (x^{4} + y^{4} - 4 x y + 2)

\int_{0}^{2} - 7 x^{3} x^{2} + 4 d x

\frac{d}{d x} ((1 + 4 x)^{\frac{3}{2}})

\int_{0}^{\infty} 2 e^{- 2 x} d x

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