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}}}

Compute the slope of

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

x = \frac{1}{16} : 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

\frac{d}{d x} (\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 6 x + 5)

f (x) = ln (sin (x^{2} + 1))

\frac{d}{d x} (x^{3} - x^{2} - x + 1)

\int cos (x + \frac{π}{2}) d x

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