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

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

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

tangent of

y = x^{3} - x^{2} + x - 1, (1, 0)

y = 2 x - 2

Solution steps

Find the slope of

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

Compute the slope of

y = x^{3} - x^{2} + x - 1

(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 x^{4} ln^{2} (x) d x

\frac{d}{d x} (\frac{4 x}{x + 4})

y = 8 + 5 x^{2} - 2 x^{3}

\frac{4 s}{( s + 2 ) ^{2}}

\frac{d}{d y} (\frac{x}{y})

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