What is the tangent of f(x)= 9/(8-x) ?

The tangent of f(x)= 9/(8-x) is y= 9/((8-a_{0))^2}x+(-18a_{0}+72)/((-a_{0)+8)^2}

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tangent of f(x)= 9/(8-x)

tangent of

f (x) = \frac{9}{8 - x}

y = \frac{9}{( 8 - a _{0} ) ^{2}} x + \frac{- 18 a _{0} + 72}{( - a _{0} + 8 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{9}{8 - a _{0}})

Find the slope of

f (x) = \frac{9}{8 - x} : \frac{df ( x )}{d x} = \frac{9}{( 8 - x ) ^{2}}

Compute the slope of

f (x) = \frac{9}{8 - x}

x = a_{0} : m = \frac{9}{( 8 - a _{0} ) ^{2}}

Find the line with slope m=

\frac{9}{( 8 - a _{0} ) ^{2}}

and passing through

(a_{0}, \frac{9}{8 - a _{0}}) : y = \frac{9}{( 8 - a _{0} ) ^{2}} x + \frac{- 18 a _{0} + 72}{( - a _{0} + 8 ) ^{2}}

y = \frac{9}{( 8 - a _{0} ) ^{2}} x + \frac{- 18 a _{0} + 72}{( - a _{0} + 8 ) ^{2}}

- 3 x^{\frac{3}{2}} - 1, at x = 4

\frac{d}{d x} (\frac{( ln ( x ) )}{x})

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

f (x) = 2 x^{2} - 3 x + 4

\int \frac{7}{( x ^{2} + 3 x - 10 )} d x

What is the tangent of f(x)= 9/(8-x) ?
The tangent of f(x)= 9/(8-x) is y= 9/((8-a_{0))^2}x+(-18a_{0}+72)/((-a_{0)+8)^2}