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

The tangent of f(x)=(1-x)(x^2-3)2 is y=2(-3a_{0}^2+2a_{0}+3)x+4a_{0}^3-2a_{0}^2-6

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

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

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

and passing through

(a_{0}, (1 - a_{0}) (a_{0}^{2} - 3) 2) : y = 2 (- 3 a_{0}^{2} + 2 a_{0} + 3) x + 4 a_{0}^{3} - 2 a_{0}^{2} - 6

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

x \to - 1 lim (x^{3})

\frac{\partial}{\partial x} (7 y)

\int_{1}^{5} (6 - x) - (8 - 3 x) d x

\frac{\partial}{\partial y} (x e^{- y})

y = x^{3} + e^{x}

What is the tangent of f(x)=(1-x)(x^2-3)2 ?
The tangent of f(x)=(1-x)(x^2-3)2 is y=2(-3a_{0}^2+2a_{0}+3)x+4a_{0}^3-2a_{0}^2-6