What is the tangent of f(x)=x^2-1/x ,(1,0) ?

The tangent of f(x)=x^2-1/x ,(1,0) is y=3x-3

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tangent of f(x)=x^2-1/x ,(1,0)

tangent of

f (x) = x^{2} - \frac{1}{x}, (1, 0)

y = 3 x - 3

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 0) : m = 3

Find the line with slope m=

3

and passing through

(1, 0) : y = 3 x - 3

y = 3 x - 3

\frac{d}{d x} (\frac{1}{9 - x ^{2} - y ^{2} - z ^{2}})

- 2 x e^{- x^{2}}

\frac{\partial}{\partial x} (x^{2} y + x + y)

((x - 1)^{2})^{^{'}}

\frac{\partial}{\partial z} (7 z e^{x yz})

What is the tangent of f(x)=x^2-1/x ,(1,0) ?
The tangent of f(x)=x^2-1/x ,(1,0) is y=3x-3