What is the tangent of f(x)=sqrt(x-4),(13,3) ?

The tangent of f(x)=sqrt(x-4),(13,3) is y= 1/6 x+5/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=sqrt(x-4),(13,3)

tangent of

f (x) = x - 4, (13, 3)

y = \frac{1}{6} x + \frac{5}{6}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x - 4

(13, 3) : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(13, 3) : y = \frac{1}{6} x + \frac{5}{6}

y = \frac{1}{6} x + \frac{5}{6}

f (x) = \frac{5}{x + 1}, at x = \frac{- 1}{3}

x \to 0 lim (\frac{0 + 5}{3 ( 0 )})

f (x) = \frac{x ^{2} + 8 x + 3}{x}

x \to \infty lim (\frac{x ^{2} + 2 x + 5}{e ^{3 x}})

\frac{d}{d x} (e^{x} cos (x) + e^{x} sin (x))

What is the tangent of f(x)=sqrt(x-4),(13,3) ?
The tangent of f(x)=sqrt(x-4),(13,3) is y= 1/6 x+5/6