What is the tangent of f(x)= 3/(sqrt(x)),\at x= 1/16 ?

The tangent of f(x)= 3/(sqrt(x)),\at x= 1/16 is y=-96x+18

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tangent of f(x)= 3/(sqrt(x)),\at x= 1/16

tangent of

f (x) = \frac{3}{x}, at x = \frac{1}{16}

y = - 96 x + 18

Solution steps

Find the tangent point:

(\frac{1}{16}, 12)

Find the slope of

f (x) = \frac{3}{x} : \frac{df ( x )}{d x} = - \frac{3}{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 = - 96

Find the line with slope m=

- 96

and passing through

(\frac{1}{16}, 12) : y = - 96 x + 18

y = - 96 x + 18

\int 2 tan^{3} (2 x) sec^{5} (2 x) d x

\int cos (x + \frac{π}{2}) d x

x \to π + lim (h (x))

\frac{d}{d x} ((7 x^{6} + 8 x^{3})^{4})

y^{^{'}} = x - y

What is the tangent of f(x)= 3/(sqrt(x)),\at x= 1/16 ?
The tangent of f(x)= 3/(sqrt(x)),\at x= 1/16 is y=-96x+18