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

The tangent of f(x)=-4x^4+x^3 is y=(-16a_{0}^3+3a_{0}^2)x+12a_{0}^4-2a_{0}^3

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tangent of f(x)=-4x^4+x^3

tangent of

f (x) = - 4 x^{4} + x^{3}

y = (- 16 a_{0}^{3} + 3 a_{0}^{2}) x + 12 a_{0}^{4} - 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 4 a_{0}^{4} + a_{0}^{3})

Find the slope of

f (x) = - 4 x^{4} + x^{3} : \frac{df ( x )}{d x} = - 16 x^{3} + 3 x^{2}

Compute the slope of

f (x) = - 4 x^{4} + x^{3}

x = a_{0} : m = - 16 a_{0}^{3} + 3 a_{0}^{2}

Find the line with slope m=

- 16 a_{0}^{3} + 3 a_{0}^{2}

and passing through

(a_{0}, - 4 a_{0}^{4} + a_{0}^{3}) : y = (- 16 a_{0}^{3} + 3 a_{0}^{2}) x + 12 a_{0}^{4} - 2 a_{0}^{3}

y = (- 16 a_{0}^{3} + 3 a_{0}^{2}) x + 12 a_{0}^{4} - 2 a_{0}^{3}

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

x \to π + lim (h (x))

\frac{\partial}{\partial x} (4 π^{2} L T^{- 2})

\frac{d}{d x} ((7 x^{6} + 8 x^{3})^{4})

\int x^{4} (x + 1)^{4} (2 x + 1) d x

What is the tangent of f(x)=-4x^4+x^3 ?
The tangent of f(x)=-4x^4+x^3 is y=(-16a_{0}^3+3a_{0}^2)x+12a_{0}^4-2a_{0}^3