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

The tangent of f(x)=(-1)/(sqrt(x)),\at x=16 is y= 1/128 x-3/8

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

tangent of

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

y = \frac{1}{128} x - \frac{3}{8}

Solution steps

Find the tangent point:

(16, - \frac{1}{4})

Find the slope of

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

Compute the slope of

f (x) = \frac{- 1}{x}

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

Find the line with slope m=

\frac{1}{128}

and passing through

(16, - \frac{1}{4}) : y = \frac{1}{128} x - \frac{3}{8}

y = \frac{1}{128} x - \frac{3}{8}

\int sin (6 x) - 7 sin (7 x) d x

\frac{6}{x}, 3 x, \frac{x}{3}

\frac{d}{d x} (sec^{3} (2 x))

\int_{- 7}^{- 5} \frac{3}{( x + 6 ) ^{\frac{8}{11}}} d x

f (x) = 1 - tan (x), at x = - \frac{π}{4}

What is the tangent of f(x)=(-1)/(sqrt(x)),\at x=16 ?
The tangent of f(x)=(-1)/(sqrt(x)),\at x=16 is y= 1/128 x-3/8