What is the tangent of 10/3 x^3-5x^2-11x+6 ?

The tangent of 10/3 x^3-5x^2-11x+6 is y=(10a_{0}^2-10a_{0}-11)x-20/3 a_{0}^3+5a_{0}^2+6

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tangent of 10/3 x^3-5x^2-11x+6

tangent of

\frac{10}{3} x^{3} - 5 x^{2} - 11 x + 6

y = (10 a_{0}^{2} - 10 a_{0} - 11) x - \frac{20}{3} a_{0}^{3} + 5 a_{0}^{2} + 6

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{10}{3} a_{0}^{3} - 5 a_{0}^{2} - 11 a_{0} + 6)

Find the slope of

y = \frac{10}{3} x^{3} - 5 x^{2} - 11 x + 6 : \frac{d y}{d x} = 10 x^{2} - 10 x - 11

Compute the slope of

y = \frac{10}{3} x^{3} - 5 x^{2} - 11 x + 6

x = a_{0} : m = 10 a_{0}^{2} - 10 a_{0} - 11

Find the line with slope m=

10 a_{0}^{2} - 10 a_{0} - 11

and passing through

(a_{0}, \frac{10}{3} a_{0}^{3} - 5 a_{0}^{2} - 11 a_{0} + 6) : y = (10 a_{0}^{2} - 10 a_{0} - 11) x - \frac{20}{3} a_{0}^{3} + 5 a_{0}^{2} + 6

y = (10 a_{0}^{2} - 10 a_{0} - 11) x - \frac{20}{3} a_{0}^{3} + 5 a_{0}^{2} + 6

\int \frac{x}{100 x ^{2} - 1} d x

\frac{d y}{d x}, x = y^{18}

g (x) = 1 - x

y^{^{''}} + y = 0, y (0) = 1, y^{^{'}} (0) = 0

x y^{^{''}} + y^{^{'}} = 20 x

What is the tangent of 10/3 x^3-5x^2-11x+6 ?
The tangent of 10/3 x^3-5x^2-11x+6 is y=(10a_{0}^2-10a_{0}-11)x-20/3 a_{0}^3+5a_{0}^2+6