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

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

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

tangent of

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

y = 2 x + 2

Solution steps

Find the tangent point:

(- 1, 0)

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

x = - 1 : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x + 2

x \to \infty lim (5^{x} + 5)

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

\frac{d}{d y} (y^{3} + k x y^{4} - 2 x)

\int \frac{1}{x ^{2}} sin (\frac{1}{x}) cos (\frac{1}{x}) d x

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