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

The tangent of y=2x^3-3x,(1,-1) is y=3x-4

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

tangent of

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

y = 3 x - 4

Solution steps

Find the slope of

y = 2 x^{3} - 3 x : \frac{d y}{d x} = 6 x^{2} - 3

Compute the slope of

y = 2 x^{3} - 3 x

(1, - 1) : m = 3

Find the line with slope m=

3

and passing through

(1, - 1) : y = 3 x - 4

y = 3 x - 4

4 x y (\frac{d y}{d x}) = 3

x \to 0 lim (b + 0^{c} \cdot ln (0))

k \to \infty lim (2 k + 1)

\int_{0}^{4} (16 - x^{2}) d x

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

What is the tangent of y=2x^3-3x,(1,-1) ?
The tangent of y=2x^3-3x,(1,-1) is y=3x-4