What is the tangent of f(x)= 4/(sqrt(x)),\at x= 1/16 ?

The tangent of f(x)= 4/(sqrt(x)),\at x= 1/16 is y=-128x+24

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tangent of f(x)= 4/(sqrt(x)),\at x= 1/16

tangent of

f (x) = \frac{4}{x}, at x = \frac{1}{16}

y = - 128 x + 24

Solution steps

Find the tangent point:

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

Find the slope of

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

Compute the slope of

f (x) = \frac{4}{x}

x = \frac{1}{16} : m = - 128

Find the line with slope m=

- 128

and passing through

(\frac{1}{16}, 16) : y = - 128 x + 24

y = - 128 x + 24

\int (3 y - 2 y - 3) d y

\frac{d}{d x} ((8 x - 7)^{- 5})

x \to 0 lim (1 + x - 1)

y \to 0 lim (\frac{sin ( 5 y )}{7 y})

\int x lo g_{e} (x) d x

What is the tangent of f(x)= 4/(sqrt(x)),\at x= 1/16 ?
The tangent of f(x)= 4/(sqrt(x)),\at x= 1/16 is y=-128x+24