What is the tangent of h(x)= 1/(2sqrt(x)),\at x=4 ?

The tangent of h(x)= 1/(2sqrt(x)),\at x=4 is y=-1/32 x+3/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of h(x)= 1/(2sqrt(x)),\at x=4

tangent of

h (x) = \frac{1}{2 x}, at x = 4

y = - \frac{1}{32} x + \frac{3}{8}

Solution steps

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

x = 4 : m = - \frac{1}{32}

Find the line with slope m=

- \frac{1}{32}

and passing through

(4, \frac{1}{4}) : y = - \frac{1}{32} x + \frac{3}{8}

y = - \frac{1}{32} x + \frac{3}{8}

x \to - 2 lim (2 x^{2})

\int x^{3} 5 + x^{2} d x

x + 7

\int_{0}^{4} 1 + 36 x d x

\int_{0}^{π} sin (x) cos (2 x) d x

What is the tangent of h(x)= 1/(2sqrt(x)),\at x=4 ?
The tangent of h(x)= 1/(2sqrt(x)),\at x=4 is y=-1/32 x+3/8