What is the tangent of f(x)=2x^3+3x^2-12x+1 ?

The tangent of f(x)=2x^3+3x^2-12x+1 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+1

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tangent of f(x)=2x^3+3x^2-12x+1

tangent of

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 1

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 1

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} + 1)

Find the slope of

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 1 : \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} + 1) : y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 1

y = (6 a_{0}^{2} + 6 a_{0} - 12) x - 4 a_{0}^{3} - 3 a_{0}^{2} + 1

\frac{d ^{2} y}{d x ^{2}} = 4 x cos (x^{2})

3 y^{^{'}} + y = (1 - 2 x) y^{4}

y = \frac{- 1}{3 x ^{2}}

\frac{d}{d x} (3 x - 4 x^{3} + 2^{x} - e + 7)

y = 1 + x^{3}, (2, 3)

What is the tangent of f(x)=2x^3+3x^2-12x+1 ?
The tangent of f(x)=2x^3+3x^2-12x+1 is y=(6a_{0}^2+6a_{0}-12)x-4a_{0}^3-3a_{0}^2+1