What is the tangent of x^3-3x+1(3.19) ?

The tangent of x^3-3x+1(3.19) is y=(3a_{0}^2-3)x-2a_{0}^3+3.19

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tangent of x^3-3x+1(3.19)

tangent of

x^{3} - 3 x + 1 (3.19)

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 3.19

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 3 a_{0} + 3.19)

Find the slope of

y = x^{3} - 3 x + 1 \cdot 3.19 : \frac{d y}{d x} = 3 x^{2} - 3

Compute the slope of

y = x^{3} - 3 x + 1 \cdot 3.19

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

Find the line with slope m=

3 a_{0}^{2} - 3

and passing through

(a_{0}, a_{0}^{3} - 3 a_{0} + 3.19) : y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 3.19

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 3.19

\int tan^{8} (x) sec^{2} (x) d x

x y^{^{'}} = - y - x, y (1) = 1

x \to - 2 lim (\frac{- x}{- x + 2})

(7 x)^{x}

\int \frac{2 x + 5}{x ^{2} + x} d x

What is the tangent of x^3-3x+1(3.19) ?
The tangent of x^3-3x+1(3.19) is y=(3a_{0}^2-3)x-2a_{0}^3+3.19