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

n = 0 \sum \infty \frac{1}{n ( n + 8 )}

\frac{5}{s \cdot ( 2 \cdot s + 1 )}

3 y^{^{'}} - 6 y = 12, y (0) = 4

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

\int x (- 4 - 7 x^{4}) d x

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