What is the tangent of f(x)=(x^3)/2-4/(3x),\at x=2 ?

The tangent of f(x)=(x^3)/2-4/(3x),\at x=2 is y= 19/3 x-28/3

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tangent of f(x)=(x^3)/2-4/(3x),\at x=2

tangent of

f (x) = \frac{x ^{3}}{2} - \frac{4}{3 x}, at x = 2

y = \frac{19}{3} x - \frac{28}{3}

Solution steps

Find the tangent point:

(2, \frac{10}{3})

Find the slope of

f (x) = \frac{x ^{3}}{2} - \frac{4}{3 x} : \frac{df ( x )}{d x} = \frac{3 x ^{2}}{2} + \frac{4}{3 x ^{2}}

Compute the slope of

f (x) = \frac{x ^{3}}{2} - \frac{4}{3 x}

x = 2 : m = \frac{19}{3}

Find the line with slope m=

\frac{19}{3}

and passing through

(2, \frac{10}{3}) : y = \frac{19}{3} x - \frac{28}{3}

y = \frac{19}{3} x - \frac{28}{3}

y = e^{m x} m

\frac{d}{d x} (3 x^{2} - 3 x)

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

(a^{3})^{^{'}}

x \to 4 lim (6 - 3 x)

What is the tangent of f(x)=(x^3)/2-4/(3x),\at x=2 ?
The tangent of f(x)=(x^3)/2-4/(3x),\at x=2 is y= 19/3 x-28/3