What is the tangent of f(x)= 1/(sqrt(3x)),\at x=9 ?

The tangent of f(x)= 1/(sqrt(3x)),\at x=9 is y=-1/(54sqrt(3))x+1/(2sqrt(3))

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tangent of f(x)= 1/(sqrt(3x)),\at x=9

tangent of

f (x) = \frac{1}{3 x}, at x = 9

y = - \frac{1}{54 3} x + \frac{1}{2 3}

Solution steps

Find the tangent point:

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

Find the slope of

f (x) = \frac{1}{3 x} : \frac{df ( x )}{d x} = - \frac{1}{2 3 x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{1}{3 x}

x = 9 : m = - \frac{1}{54 3}

Find the line with slope m=

- \frac{1}{54 3}

and passing through

(9, \frac{1}{3 3}) : y = - \frac{1}{54 3} x + \frac{1}{2 3}

y = - \frac{1}{54 3} x + \frac{1}{2 3}

x \to - 7 lim (\frac{7 - ∣ x ∣}{7 + x})

\int_{1}^{5} \frac{ln ( y )}{x y} d y

(e^{- 4 x})^{^{'}}

\int 30 x^{2} d x

x \to 9 lim (x - 9)

What is the tangent of f(x)= 1/(sqrt(3x)),\at x=9 ?
The tangent of f(x)= 1/(sqrt(3x)),\at x=9 is y=-1/(54sqrt(3))x+1/(2sqrt(3))