What is the tangent of y=x^3-2x+1 ?

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

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tangent of y=x^3-2x+1

tangent of

y = x^{3} - 2 x + 1

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = x^{3} - 2 x + 1

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

Find the line with slope m=

3 a_{0}^{2} - 2

and passing through

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

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

x \to 0 lim (\frac{∣ x ( h + 1 ) ∣}{x})

\int_{0}^{8} \frac{x}{x ^{2} - 1} d x

\frac{\partial}{\partial y} (x^{4} + y^{4} - 4 x y + 2)

x + \frac{5}{x}

\int_{0}^{1} \frac{10}{4 y - 1} d y

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