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

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

x 3 x - 4

n = 1 \sum \infty 2 (- e)^{- n}

\int \frac{x ^{5}}{x ^{2} + 7} d x

\int \frac{1}{5 + 4 x - x ^{2}} d x

x \to - \infty lim (\frac{( 4 x ^{2} + 1 )}{( 2 - 7 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