What is the tangent of y=x^3-16x+2 ?

The tangent of y=x^3-16x+2 is y=(3a_{0}^2-16)x-2a_{0}^3+2

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tangent of y=x^3-16x+2

tangent of

y = x^{3} - 16 x + 2

y = (3 a_{0}^{2} - 16) 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} - 16 a_{0} + 2)

Find the slope of

y = x^{3} - 16 x + 2 : \frac{d y}{d x} = 3 x^{2} - 16

Compute the slope of

y = x^{3} - 16 x + 2

x = a_{0} : m = 3 a_{0}^{2} - 16

Find the line with slope m=

3 a_{0}^{2} - 16

and passing through

(a_{0}, a_{0}^{3} - 16 a_{0} + 2) : y = (3 a_{0}^{2} - 16) x - 2 a_{0}^{3} + 2

y = (3 a_{0}^{2} - 16) x - 2 a_{0}^{3} + 2

\int r^{- 2} d r

x \to 0 + lim (\frac{e ^{x}}{ln ( x )})

\int x^{2} + 2 x - 3 d x

\int ln (t) t^{13} d t

\int (x^{2} + 4) d x

What is the tangent of y=x^3-16x+2 ?
The tangent of y=x^3-16x+2 is y=(3a_{0}^2-16)x-2a_{0}^3+2