What is the tangent of (x^3+1)/(2x) ?

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

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tangent of (x^3+1)/(2x)

tangent of

\frac{x ^{3} + 1}{2 x}

y = \frac{2 a _{0}^{3} - 1}{2 a _{0}^{2}} x + \frac{- a _{0}^{3} + 2}{2 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0}^{3} + 1}{2 a _{0}})

Find the slope of

y = \frac{x ^{3} + 1}{2 x} : \frac{d y}{d x} = \frac{2 x ^{3} - 1}{2 x ^{2}}

Compute the slope of

y = \frac{x ^{3} + 1}{2 x}

x = a_{0} : m = \frac{2 a _{0}^{3} - 1}{2 a _{0}^{2}}

Find the line with slope m=

\frac{2 a _{0}^{3} - 1}{2 a _{0}^{2}}

and passing through

(a_{0}, \frac{a _{0}^{3} + 1}{2 a _{0}}) : y = \frac{2 a _{0}^{3} - 1}{2 a _{0}^{2}} x + \frac{- a _{0}^{3} + 2}{2 a _{0}}

y = \frac{2 a _{0}^{3} - 1}{2 a _{0}^{2}} x + \frac{- a _{0}^{3} + 2}{2 a _{0}}

x \to 8 lim (\frac{x + 6}{x - 8})

\int 16 x e^{- x^{2}} d x

\frac{ln ( u )}{u}

\int \frac{1}{( x ^{2} - 4 x + 3 ) ( x ^{2} + 4 x + 5 )} d x

x \to - 2 lim (- x^{2} + 6 x - 5)

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