What is the tangent of y=x^3-2x+2(2.6) ?

The tangent of y=x^3-2x+2(2.6) is y=(3a_{0}^2-2)x-2a_{0}^3+5.2

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tangent of y=x^3-2x+2(2.6)

tangent of

y = x^{3} - 2 x + 2 (2.6)

y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 5.2

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} + 5.2)

Find the slope of

y = x^{3} - 2 x + 2 \cdot 2.6 : \frac{d y}{d x} = 3 x^{2} - 2

Compute the slope of

y = x^{3} - 2 x + 2 \cdot 2.6

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} + 5.2) : y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 5.2

y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 5.2

\frac{d}{d x} (x + \frac{2}{sin ( x )})

\int_{0}^{\frac{π}{2}} 24 x^{2} sin (x) d x

x \to 0 - lim (\frac{3}{x ^{2} - x})

\int \frac{- x - 8}{x ^{2} + 9} d x

2 x^{2} - 2 x

What is the tangent of y=x^3-2x+2(2.6) ?
The tangent of y=x^3-2x+2(2.6) is y=(3a_{0}^2-2)x-2a_{0}^3+5.2