What is the tangent of f(x)=12sqrt(x),\at x=1 ?

The tangent of f(x)=12sqrt(x),\at x=1 is y=6x+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)=12sqrt(x),\at x=1

tangent of

f (x) = 12 x, at x = 1

y = 6 x + 6

Solution steps

Find the tangent point:

(1, 12)

Find the slope of

f (x) = 12 x : \frac{df ( x )}{d x} = \frac{6}{x}

Compute the slope of

f (x) = 12 x

x = 1 : m = 6

Find the line with slope m=

6

and passing through

(1, 12) : y = 6 x + 6

y = 6 x + 6

\frac{\partial}{\partial y} (\frac{x y z ^{2}}{x y + 1})

f (x) = 6 x, (6, 6)

\frac{30 s + 102}{s ^{2} + 4 s + 40}

\int_{4}^{\infty} e^{- 4 x} d x

\frac{d}{d x} ((4 x - 2)^{2})

What is the tangent of f(x)=12sqrt(x),\at x=1 ?
The tangent of f(x)=12sqrt(x),\at x=1 is y=6x+6