What is the tangent of f(x)= 1/(2sqrt(x)),\at x=64 ?

The tangent of f(x)= 1/(2sqrt(x)),\at x=64 is y=-1/2048 x+3/32

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)= 1/(2sqrt(x)),\at x=64

tangent of

f (x) = \frac{1}{2 x}, at x = 64

y = - \frac{1}{2048} x + \frac{3}{32}

Solution steps

Find the tangent point:

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

Find the slope of

f (x) = \frac{1}{2 x} : \frac{df ( x )}{d x} = - \frac{1}{4 x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{1}{2 x}

x = 64 : m = - \frac{1}{2048}

Find the line with slope m=

- \frac{1}{2048}

and passing through

(64, \frac{1}{16}) : y = - \frac{1}{2048} x + \frac{3}{32}

y = - \frac{1}{2048} x + \frac{3}{32}

\frac{d}{d x} (cos (x + π))

\frac{d}{d x} (x^{2} + 4 x - 5)

\int sec^{3} (4 x) tan (4 x) d x

\int \frac{1}{5 x} d x

\int_{0}^{2} 2 x^{2} e^{- 2 x} d x

What is the tangent of f(x)= 1/(2sqrt(x)),\at x=64 ?
The tangent of f(x)= 1/(2sqrt(x)),\at x=64 is y=-1/2048 x+3/32