What is the tangent of f(x)=x^3-9x(3) ?

The tangent of f(x)=x^3-9x(3) is y=(3a_{0}^2-27)x-2a_{0}^3

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tangent of f(x)=x^3-9x(3)

tangent of

f (x) = x^{3} - 9 x (3)

y = (3 a_{0}^{2} - 27) x - 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 27 a_{0})

Find the slope of

f (x) = x^{3} - 9 x 3 : \frac{df ( x )}{d x} = 3 x^{2} - 27

Compute the slope of

f (x) = x^{3} - 9 x 3

x = a_{0} : m = 3 a_{0}^{2} - 27

Find the line with slope m=

3 a_{0}^{2} - 27

and passing through

(a_{0}, a_{0}^{3} - 27 a_{0}) : y = (3 a_{0}^{2} - 27) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} - 27) x - 2 a_{0}^{3}

f (x) = tan^{2} (x), at x = \frac{- π}{4}

\frac{\partial}{\partial x} ((2 x - 5 y)^{2})

\int y^{3} + 10 x y^{4} - 2 x d x

x \to - 4 - lim (\frac{x + 3}{x + 4})

f (x) = x 8 - x^{2}

What is the tangent of f(x)=x^3-9x(3) ?
The tangent of f(x)=x^3-9x(3) is y=(3a_{0}^2-27)x-2a_{0}^3