What is the tangent of f(x)=x^3-4x^2-1 ?

The tangent of f(x)=x^3-4x^2-1 is y=(3a_{0}^2-8a_{0})x-2a_{0}^3+4a_{0}^2-1

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tangent of f(x)=x^3-4x^2-1

tangent of

f (x) = x^{3} - 4 x^{2} - 1

y = (3 a_{0}^{2} - 8 a_{0}) x - 2 a_{0}^{3} + 4 a_{0}^{2} - 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 4 a_{0}^{2} - 1)

Find the slope of

f (x) = x^{3} - 4 x^{2} - 1 : \frac{df ( x )}{d x} = 3 x^{2} - 8 x

Compute the slope of

f (x) = x^{3} - 4 x^{2} - 1

x = a_{0} : m = 3 a_{0}^{2} - 8 a_{0}

Find the line with slope m=

3 a_{0}^{2} - 8 a_{0}

and passing through

(a_{0}, a_{0}^{3} - 4 a_{0}^{2} - 1) : y = (3 a_{0}^{2} - 8 a_{0}) x - 2 a_{0}^{3} + 4 a_{0}^{2} - 1

y = (3 a_{0}^{2} - 8 a_{0}) x - 2 a_{0}^{3} + 4 a_{0}^{2} - 1

\frac{d y}{d t} = - \frac{y}{t} - 8

\frac{d}{d x} (e^{4 x} cos (3 x))

\int sin (4 x) cos (4 x) d x

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

What is the tangent of f(x)=x^3-4x^2-1 ?
The tangent of f(x)=x^3-4x^2-1 is y=(3a_{0}^2-8a_{0})x-2a_{0}^3+4a_{0}^2-1