What is the tangent of g(x)= 1/(2sqrt(x)),\at x=9 ?

The tangent of g(x)= 1/(2sqrt(x)),\at x=9 is y=-1/108 x+1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of g(x)= 1/(2sqrt(x)),\at x=9

tangent of

g (x) = \frac{1}{2 x}, at x = 9

y = - \frac{1}{108} x + \frac{1}{4}

Solution steps

Find the tangent point:

(9, \frac{1}{6})

Find the slope of

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

Compute the slope of

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

x = 9 : m = - \frac{1}{108}

Find the line with slope m=

- \frac{1}{108}

and passing through

(9, \frac{1}{6}) : y = - \frac{1}{108} x + \frac{1}{4}

y = - \frac{1}{108} x + \frac{1}{4}

x^{\frac{1}{2}}, x - 2, [0, 4]

7 x - 4 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 6

7 tan (x)

\frac{d}{d x} (e x^{2} - y)

\int x 1 - x d x

What is the tangent of g(x)= 1/(2sqrt(x)),\at x=9 ?
The tangent of g(x)= 1/(2sqrt(x)),\at x=9 is y=-1/108 x+1/4