What is the tangent of y=x^3-2x+2(4.58) ?

The tangent of y=x^3-2x+2(4.58) is y=(3a_{0}^2-2)x-2a_{0}^3+9.16

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tangent of y=x^3-2x+2(4.58)

tangent of

y = x^{3} - 2 x + 2 (4.58)

y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 9.16

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} + 9.16)

Find the slope of

y = x^{3} - 2 x + 2 \cdot 4.58 : \frac{d y}{d x} = 3 x^{2} - 2

Compute the slope of

y = x^{3} - 2 x + 2 \cdot 4.58

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} + 9.16) : y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 9.16

y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 9.16

x^{3} + x^{2} - 6 x, - x^{2} + 2 x

\frac{d}{d x} (5 x^{- 2} + 500)

\int x^{5} I n (x) d x

\frac{d ^{2}}{d x ^{2}} (8 x^{- 2})

What is the tangent of y=x^3-2x+2(4.58) ?
The tangent of y=x^3-2x+2(4.58) is y=(3a_{0}^2-2)x-2a_{0}^3+9.16