What is the tangent of f(x)= 1/(sqrt(8x)),\at x=9 ?

The tangent of f(x)= 1/(sqrt(8x)),\at x=9 is y=-1/(108sqrt(2))x+1/(4sqrt(2))

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tangent of f(x)= 1/(sqrt(8x)),\at x=9

tangent of

f (x) = \frac{1}{8 x}, at x = 9

y = - \frac{1}{108 2} x + \frac{1}{4 2}

Solution steps

Find the tangent point:

(9, \frac{1}{6 2})

Find the slope of

f (x) = \frac{1}{8 x} : \frac{df ( x )}{d x} = - \frac{1}{4 2 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 2}

Find the line with slope m=

- \frac{1}{108 2}

and passing through

(9, \frac{1}{6 2}) : y = - \frac{1}{108 2} x + \frac{1}{4 2}

y = - \frac{1}{108 2} x + \frac{1}{4 2}

What is the tangent of f(x)= 1/(sqrt(8x)),\at x=9 ?
The tangent of f(x)= 1/(sqrt(8x)),\at x=9 is y=-1/(108sqrt(2))x+1/(4sqrt(2))