What is the tangent of f(x)=sqrt(3x+1),\at x=8 ?

The tangent of f(x)=sqrt(3x+1),\at x=8 is y= 3/10 x+13/5

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(3x+1),\at x=8

tangent of

f (x) = 3 x + 1, at x = 8

y = \frac{3}{10} x + \frac{13}{5}

Solution steps

Find the tangent point:

(8, 5)

Find the slope of

f (x) = 3 x + 1 : \frac{df ( x )}{d x} = \frac{3}{2 3 x + 1}

Compute the slope of

f (x) = 3 x + 1

x = 8 : m = \frac{3}{10}

Find the line with slope m=

\frac{3}{10}

and passing through

(8, 5) : y = \frac{3}{10} x + \frac{13}{5}

y = \frac{3}{10} x + \frac{13}{5}

\int 3 1 - x^{2} d x

p (t) = 25 e^{0.03 t}

\int \frac{( x ^{3} - 1 )}{( 4 x ^{3} - x )} d x

\int \frac{- 4 x - 6}{x ^{2} + x + 1} d x

f (x) = \frac{e ^{x} x - e ^{x}}{5 x ^{2}}

What is the tangent of f(x)=sqrt(3x+1),\at x=8 ?
The tangent of f(x)=sqrt(3x+1),\at x=8 is y= 3/10 x+13/5