What is the tangent of f(x)=3x^3+3x^2-11x-1 ?

The tangent of f(x)=3x^3+3x^2-11x-1 is y=(9a_{0}^2+6a_{0}-11)x-6a_{0}^3-3a_{0}^2-1

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tangent of f(x)=3x^3+3x^2-11x-1

tangent of

f (x) = 3 x^{3} + 3 x^{2} - 11 x - 1

y = (9 a_{0}^{2} + 6 a_{0} - 11) x - 6 a_{0}^{3} - 3 a_{0}^{2} - 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{3} + 3 a_{0}^{2} - 11 a_{0} - 1)

Find the slope of

f (x) = 3 x^{3} + 3 x^{2} - 11 x - 1 : \frac{df ( x )}{d x} = 9 x^{2} + 6 x - 11

Compute the slope of

f (x) = 3 x^{3} + 3 x^{2} - 11 x - 1

x = a_{0} : m = 9 a_{0}^{2} + 6 a_{0} - 11

Find the line with slope m=

9 a_{0}^{2} + 6 a_{0} - 11

and passing through

(a_{0}, 3 a_{0}^{3} + 3 a_{0}^{2} - 11 a_{0} - 1) : y = (9 a_{0}^{2} + 6 a_{0} - 11) x - 6 a_{0}^{3} - 3 a_{0}^{2} - 1

y = (9 a_{0}^{2} + 6 a_{0} - 11) x - 6 a_{0}^{3} - 3 a_{0}^{2} - 1

y^{^{''}} + 9 y = 3 csc^{2} (3 t)

\int sinh (2 x) d x

2 x^{\frac{1}{5}} (x^{3} - 8)

x \to 0 lim (\frac{x}{cos ( x )})

\int t^{2} + t t - 1 + t sin (π) t d t

What is the tangent of f(x)=3x^3+3x^2-11x-1 ?
The tangent of f(x)=3x^3+3x^2-11x-1 is y=(9a_{0}^2+6a_{0}-11)x-6a_{0}^3-3a_{0}^2-1