What is the tangent of f(x)=sqrt(x+8),\at x=1 ?

The tangent of f(x)=sqrt(x+8),\at x=1 is y= 1/6 x+17/6

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+8),\at x=1

tangent of

f (x) = x + 8, at x = 1

y = \frac{1}{6} x + \frac{17}{6}

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

f (x) = x + 8 : \frac{df ( x )}{d x} = \frac{1}{2 x + 8}

Compute the slope of

f (x) = x + 8

x = 1 : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(1, 3) : y = \frac{1}{6} x + \frac{17}{6}

y = \frac{1}{6} x + \frac{17}{6}

x \to 1 lim (2^{x^{3} - x})

x \to 7 + lim (\frac{( x - 8 )}{x - 7})

\int \frac{1}{1 + 25 x ^{2}} d x

\frac{\partial}{\partial z} (z (1 + e^{x + y}))

\frac{d}{d x} (3 x y^{2} + 2 y cos (x))

What is the tangent of f(x)=sqrt(x+8),\at x=1 ?
The tangent of f(x)=sqrt(x+8),\at x=1 is y= 1/6 x+17/6