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

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

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tangent of x^3-2x+2

tangent of

x^{3} - 2 x + 2

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

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} + 2)

Find the slope of

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

Compute the slope of

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

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

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

\int sin (4 x) cos (17 x) d x

n = 0 \sum \infty (\frac{2}{11})^{n}

- 6 x^{2} + 5 x

y = (ln (x))^{3}

\int \frac{x + 3}{x - 1} d x

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