What is the tangent of f(x)=x^3-x,(-1,0) ?

The tangent of f(x)=x^3-x,(-1,0) is y=2x+2

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tangent of f(x)=x^3-x,(-1,0)

tangent of

f (x) = x^{3} - x, (- 1, 0)

y = 2 x + 2

Solution steps

Find the slope of

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

Compute the slope of

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

(- 1, 0) : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x + 2

\int \frac{1}{3 x + 8} d x

\int x^{2} + 16 d x

g (x) = (\frac{3 x ^{2} - 2}{2 x + 3})^{3}

5 x^{2} - 2 x

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{8} e^{- x} d y = 0

What is the tangent of f(x)=x^3-x,(-1,0) ?
The tangent of f(x)=x^3-x,(-1,0) is y=2x+2