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

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

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

tangent of

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

y = 2 x + 2

Solution steps

Find the slope of

y = x^{3} - x : \frac{d y}{d x} = 3 x^{2} - 1

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 = 2

Find the line with slope m=

2

and passing through

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

y = 2 x + 2

f (x) = \frac{2}{x}, at x = - 1

\frac{1}{x} d y - \frac{y}{x ^{2}} d x = 0

n = 1 \sum \infty (\frac{2}{7})^{n}

y^{^{'}} = \frac{- 2 x}{y}

\frac{d}{d x} (- 2 x^{5} + \frac{1}{2 x ^{2}})