What is the tangent of 3x^3-6x^2+4x+21 ?

The tangent of 3x^3-6x^2+4x+21 is y=(9a_{0}^2-12a_{0}+4)x-6a_{0}^3+6a_{0}^2+21

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tangent of 3x^3-6x^2+4x+21

tangent of

3 x^{3} - 6 x^{2} + 4 x + 21

y = (9 a_{0}^{2} - 12 a_{0} + 4) x - 6 a_{0}^{3} + 6 a_{0}^{2} + 21

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{3} - 6 a_{0}^{2} + 4 a_{0} + 21)

Find the slope of

y = 3 x^{3} - 6 x^{2} + 4 x + 21 : \frac{d y}{d x} = 9 x^{2} - 12 x + 4

Compute the slope of

y = 3 x^{3} - 6 x^{2} + 4 x + 21

x = a_{0} : m = 9 a_{0}^{2} - 12 a_{0} + 4

Find the line with slope m=

9 a_{0}^{2} - 12 a_{0} + 4

and passing through

(a_{0}, 3 a_{0}^{3} - 6 a_{0}^{2} + 4 a_{0} + 21) : y = (9 a_{0}^{2} - 12 a_{0} + 4) x - 6 a_{0}^{3} + 6 a_{0}^{2} + 21

y = (9 a_{0}^{2} - 12 a_{0} + 4) x - 6 a_{0}^{3} + 6 a_{0}^{2} + 21

\int \frac{7 x + 3}{( x + 4 ) ( x - 1 )} d x

x \to 1 lim (\frac{\frac{1}{x} - 1}{1 - x})

x \to 0 lim (\frac{18}{x ^{2}})

\frac{d}{d x} (\frac{9}{( x - 8 ) ^{2}})

\frac{\partial}{\partial x} (3 x^{2} - 24 y)

What is the tangent of 3x^3-6x^2+4x+21 ?
The tangent of 3x^3-6x^2+4x+21 is y=(9a_{0}^2-12a_{0}+4)x-6a_{0}^3+6a_{0}^2+21