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

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}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

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