What is the tangent of f(x)=3x^3+6x^2+19 ?

The tangent of f(x)=3x^3+6x^2+19 is y=(9a_{0}^2+12a_{0})x-6a_{0}^3-6a_{0}^2+19

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tangent of f(x)=3x^3+6x^2+19

tangent of

f (x) = 3 x^{3} + 6 x^{2} + 19

y = (9 a_{0}^{2} + 12 a_{0}) x - 6 a_{0}^{3} - 6 a_{0}^{2} + 19

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} + 19)

Find the slope of

f (x) = 3 x^{3} + 6 x^{2} + 19 : \frac{df ( x )}{d x} = 9 x^{2} + 12 x

Compute the slope of

f (x) = 3 x^{3} + 6 x^{2} + 19

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

Find the line with slope m=

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

and passing through

(a_{0}, 3 a_{0}^{3} + 6 a_{0}^{2} + 19) : y = (9 a_{0}^{2} + 12 a_{0}) x - 6 a_{0}^{3} - 6 a_{0}^{2} + 19

y = (9 a_{0}^{2} + 12 a_{0}) x - 6 a_{0}^{3} - 6 a_{0}^{2} + 19

(x + y)^{2} d x + (2 x y + x^{2} - 2) d y = 0

\int \frac{2}{x ^{2} + 5 x + 6} d x

x \to 0 lim (\frac{( tan ( a x ) )}{x})

\int cos (5 x) sec^{2} (5 x) d x

(3 x^{2} + 9) (x^{2} - 1 x)

What is the tangent of f(x)=3x^3+6x^2+19 ?
The tangent of f(x)=3x^3+6x^2+19 is y=(9a_{0}^2+12a_{0})x-6a_{0}^3-6a_{0}^2+19