What is the tangent of f(x)= 8/x ,\at x=1 ?

The tangent of f(x)= 8/x ,\at x=1 is y=-8x+16

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tangent of f(x)= 8/x ,\at x=1

tangent of

f (x) = \frac{8}{x}, at x = 1

y = - 8 x + 16

Solution steps

Find the tangent point:

(1, 8)

Find the slope of

f (x) = \frac{8}{x} : \frac{df ( x )}{d x} = - \frac{8}{x ^{2}}

Compute the slope of

f (x) = \frac{8}{x}

x = 1 : m = - 8

Find the line with slope m=

- 8

and passing through

(1, 8) : y = - 8 x + 16

y = - 8 x + 16

\frac{\partial}{\partial x} (sin (\frac{5 x}{y}))

\frac{\partial}{\partial x} ((3 x y - 1) e^{- x y})

\int \frac{2}{3 x + 1} d x

\int_{4}^{8} ln (x^{2}) d x

\int_{0}^{\frac{π}{3}} x sec^{2} (x) d x

What is the tangent of f(x)= 8/x ,\at x=1 ?
The tangent of f(x)= 8/x ,\at x=1 is y=-8x+16