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

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

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

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 3 a_{0}^{2} - 4 a_{0} - 3

Find the line with slope m=

3 a_{0}^{2} - 4 a_{0} - 3

and passing through

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

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

\int_{1}^{8} 1 d x

\int_{0}^{1} 7 x e^{x^{2}} d x

y^{^{''}} - 6 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

2 (3 - x^{2})^{7}

\frac{d}{d x} (cos (2 x))

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