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

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

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tangent of x^3-2x+1

tangent of

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 \cdot \frac{d y}{d x} + y = e^{x}

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

- 4 cos (2 t)

\frac{1}{5 t ^{2} - 6 t + 2}

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