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

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

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

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

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

and passing through

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

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

\frac{\partial}{\partial x} (1 + 2 x y)

\frac{s}{( s + 2 ) ^{2} + 1}

\int 7 x^{5} + 6 x^{4} - 3 x^{3} - 9 x + 5 d x

\int \frac{1}{( x )} d x

\frac{d r}{d θ} = \frac{r ^{2} + 8 r θ}{4 θ ^{2}}

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