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

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

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

tangent of

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

y = - x + 1

Solution steps

Find the tangent point:

(- 1, 2)

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}

x = - 1 : m = - 1

Find the line with slope m=

- 1

and passing through

(- 1, 2) : y = - x + 1

y = - x + 1

\frac{d}{d x} (e^{a r c s i n (x)})

\frac{d}{d x} (x + 6 x + 3)

tan (x) + cos^{2} (x) + sin^{2} (x) - 3

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

\int e^{- 8 x} tan (e^{- 8 x}) d x

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