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

The tangent of f(x)= 1/(sqrt(4x)),\at x=9 is y=-1/108 x+1/4

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

tangent of

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

y = - \frac{1}{108} x + \frac{1}{4}

Solution steps

Find the tangent point:

(9, \frac{1}{6})

Find the slope of

f (x) = \frac{1}{4 x} : \frac{df ( x )}{d x} = - \frac{1}{4 x ^{\frac{3}{2}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{1}{108}

Find the line with slope m=

- \frac{1}{108}

and passing through

(9, \frac{1}{6}) : y = - \frac{1}{108} x + \frac{1}{4}

y = - \frac{1}{108} x + \frac{1}{4}

\frac{d}{d x} (5 x - 32 x)

\int \frac{5000 q}{25 q + 1500} d q

\int_{0}^{1} 40 x d x

x \to 0 lim (+ (6 - \frac{1}{x ^{3}}))

n = 3 \sum \infty (\frac{8}{11})^{- n}

What is the tangent of f(x)= 1/(sqrt(4x)),\at x=9 ?
The tangent of f(x)= 1/(sqrt(4x)),\at x=9 is y=-1/108 x+1/4