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

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

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

tangent of

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

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

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} + 2)

Find the slope of

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

Compute the slope of

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

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

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

x \to 1 lim (x^{2} + 1)

2 x y d y + (x^{2} + y^{2}) d x = 0

\int (\frac{x ( x - 2 )}{( x - 1 ) ^{3}}) d x

f (x) = sinh (4 x)

\frac{t d u}{d t} = t^{2} + 3 u

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