What is the tangent of f(x)=2x^3+3x^2-12x+4 ?

The tangent of f(x)=2x^3+3x^2-12x+4 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+4

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tangent of f(x)=2x^3+3x^2-12x+4

tangent of

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} + 3 a_{0}^{2} - 12 a_{0} + 4)

Find the slope of

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4 : \frac{df ( x )}{d x} = 6 x^{2} + 6 x - 12

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 6 a_{0}^{2} + 6 a_{0} - 12

Find the line with slope m=

6 a_{0}^{2} + 6 a_{0} - 12

and passing through

(a_{0}, 2 a_{0}^{3} + 3 a_{0}^{2} - 12 a_{0} + 4) : y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 4

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 4

\frac{\partial}{\partial x} (c y^{- 3})

\int 2 - \frac{x}{2} d x

3 x^{4}

\frac{s - 2}{( s - 3 ) ( s - 1 )}

f (x) = \frac{3}{1 + x ^{2}}, (1, \frac{3}{2})

What is the tangent of f(x)=2x^3+3x^2-12x+4 ?
The tangent of f(x)=2x^3+3x^2-12x+4 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+4