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

The tangent of f(x)=3+4x^2-2x^3 is y=(-6a_{0}^2+8a_{0})x+4a_{0}^3-4a_{0}^2+3

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

tangent of

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

y = (- 6 a_{0}^{2} + 8 a_{0}) x + 4 a_{0}^{3} - 4 a_{0}^{2} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = 3 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 x

Compute the slope of

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

x = a_{0} : m = - 6 a_{0}^{2} + 8 a_{0}

Find the line with slope m=

- 6 a_{0}^{2} + 8 a_{0}

and passing through

(a_{0}, 3 + 4 a_{0}^{2} - 2 a_{0}^{3}) : y = (- 6 a_{0}^{2} + 8 a_{0}) x + 4 a_{0}^{3} - 4 a_{0}^{2} + 3

y = (- 6 a_{0}^{2} + 8 a_{0}) x + 4 a_{0}^{3} - 4 a_{0}^{2} + 3

\int \frac{1}{x ^{2} - y} d x

f (x) = tan (x)

- \frac{1}{2} cos^{2} (x)

y = x^{2} - 3 x - 4, (2, - 6)

\int_{0}^{2} 5^{x} d x

What is the tangent of f(x)=3+4x^2-2x^3 ?
The tangent of f(x)=3+4x^2-2x^3 is y=(-6a_{0}^2+8a_{0})x+4a_{0}^3-4a_{0}^2+3