What is the tangent of y=2x^3-5x,\at x=(3-2) ?

The tangent of y=2x^3-5x,\at x=(3-2) is y=x-4

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tangent of y=2x^3-5x,\at x=(3-2)

tangent of

y = 2 x^{3} - 5 x, at x = (3 - 2)

y = x - 4

Solution steps

Find the tangent point:

((3 - 2), - 3)

Find the slope of

y = 2 x^{3} - 5 x : \frac{d y}{d x} = 6 x^{2} - 5

Compute the slope of

y = 2 x^{3} - 5 x

x = (3 - 2) : m = 1

Find the line with slope m=

1

and passing through

((3 - 2), - 3) : y = x - 4

y = x - 4

\frac{d}{d x} ((3 + x - 2 x^{2})^{\frac{1}{4}})

y^{^{'}} = \frac{( 2 - y ) ^{2}}{2 \cdot 1 + x}

x = 2 y^{2}, x = 4 + y^{2}

y = x^{3}, y = 9 x

\int 2 cos (x) - sin (x) d x

What is the tangent of y=2x^3-5x,\at x=(3-2) ?
The tangent of y=2x^3-5x,\at x=(3-2) is y=x-4