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

The tangent of f(x)=-1/x ,\at x=3 is y= 1/9 x-2/3

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

tangent of

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

y = \frac{1}{9} x - \frac{2}{3}

Solution steps

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

x = 3 : m = \frac{1}{9}

Find the line with slope m=

\frac{1}{9}

and passing through

(3, - \frac{1}{3}) : y = \frac{1}{9} x - \frac{2}{3}

y = \frac{1}{9} x - \frac{2}{3}

f (x) = 2 x^{2} + 3 x, at x = - 3

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

\int e^{2 x} arcsin (e^{x}) d x

k_{p} \cdot e (t)

x \to - 4 lim (- x^{2} + 6 x - 8)

What is the tangent of f(x)=-1/x ,\at x=3 ?
The tangent of f(x)=-1/x ,\at x=3 is y= 1/9 x-2/3