What is the tangent of f(x)= 4/(sqrt(x)),\at x= 1/9 ?

The tangent of f(x)= 4/(sqrt(x)),\at x= 1/9 is y=-54x+18

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tangent of f(x)= 4/(sqrt(x)),\at x= 1/9

tangent of

f (x) = \frac{4}{x}, at x = \frac{1}{9}

y = - 54 x + 18

Solution steps

Find the tangent point:

(\frac{1}{9}, 12)

Find the slope of

f (x) = \frac{4}{x} : \frac{df ( x )}{d x} = - \frac{2}{x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{4}{x}

x = \frac{1}{9} : m = - 54

Find the line with slope m=

- 54

and passing through

(\frac{1}{9}, 12) : y = - 54 x + 18

y = - 54 x + 18

\int 16 x ln (5 x) d x

\frac{\partial}{\partial x} (x (x^{2} + y^{2})^{- 1})

f (x) = 4 x cos (x)

x - 9

\int \frac{1}{2 x ^{2} + 4 x + 7} d x

What is the tangent of f(x)= 4/(sqrt(x)),\at x= 1/9 ?
The tangent of f(x)= 4/(sqrt(x)),\at x= 1/9 is y=-54x+18