What is the tangent of f(x)=x^2-1/x ,\at (-1) ?

The tangent of f(x)=x^2-1/x ,\at (-1) is y=(2a_{0}+1/(a_{0)^2})x-2/(a_{0)}-a_{0}^2

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tangent of f(x)=x^2-1/x ,\at (-1)

tangent of

f (x) = x^{2} - \frac{1}{x}, at (- 1)

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = x^{2} - \frac{1}{x} : \frac{df ( x )}{d x} = 2 x + \frac{1}{x ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 a_{0} + \frac{1}{a _{0}^{2}}

Find the line with slope m=

2 a_{0} + \frac{1}{a _{0}^{2}}

and passing through

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

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

\int \frac{( 2 x + 7 )}{2 x ^{2} + 2 x + 1} d x

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

f (x) = \frac{7}{x}, at x = 6

y^{^{'}} = \frac{e ^{x} - e ^{y}}{1 + e ^{x}}

\int \frac{sin ( 2 x )}{12 + cos ^{2} ( x )} d x

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