What is the tangent of f(x)=(x^3)/3-3/(x^3),\at x=-1 ?

The tangent of f(x)=(x^3)/3-3/(x^3),\at x=-1 is y=10x+38/3

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tangent of f(x)=(x^3)/3-3/(x^3),\at x=-1

tangent of

f (x) = \frac{x ^{3}}{3} - \frac{3}{x ^{3}}, at x = - 1

y = 10 x + \frac{38}{3}

Solution steps

Find the tangent point:

(- 1, \frac{8}{3})

Find the slope of

f (x) = \frac{x ^{3}}{3} - \frac{3}{x ^{3}} : \frac{df ( x )}{d x} = x^{2} + \frac{9}{x ^{4}}

Compute the slope of

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

x = - 1 : m = 10

Find the line with slope m=

10

and passing through

(- 1, \frac{8}{3}) : y = 10 x + \frac{38}{3}

y = 10 x + \frac{38}{3}

f (x) = (\frac{2 x}{7} + 1)

\int x^{2} sin^{2} (x) d x

x \to 2 - lim (x + 1)

arccot (u)

5 \cdot y^{^{''}} - 4 y^{^{'}} = 0

What is the tangent of f(x)=(x^3)/3-3/(x^3),\at x=-1 ?
The tangent of f(x)=(x^3)/3-3/(x^3),\at x=-1 is y=10x+38/3