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

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

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tangent of f(x)=-x^3

tangent of

f (x) = - x^{3}

y = - 3 a_{0}^{2} 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})

Find the slope of

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

Compute the slope of

f (x) = - x^{3}

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

Find the line with slope m=

- 3 a_{0}^{2}

and passing through

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

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

\frac{\partial}{\partial y} (\frac{1}{2} sin (2 x - y))

\int_{5}^{\infty} x e^{- 2 x} d x

f (x) = 7 x^{2} + 4 x

\int 3 e^{x} + 8 x^{3} + 4 x d x

\frac{d}{d x} (cos (\frac{1}{4} x^{3}))

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