What is the tangent of f(x)=sqrt(3x+1),(5,4) ?

The tangent of f(x)=sqrt(3x+1),(5,4) is y= 3/8 x+17/8

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),(5,4)

tangent of

f (x) = 3 x + 1, (5, 4)

y = \frac{3}{8} x + \frac{17}{8}

Solution steps

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

(5, 4) : m = \frac{3}{8}

Find the line with slope m=

\frac{3}{8}

and passing through

(5, 4) : y = \frac{3}{8} x + \frac{17}{8}

y = \frac{3}{8} x + \frac{17}{8}

\frac{d}{d x} (\frac{x ^{2} + 1}{x - 1})

t^{2} y^{^{''}} + 2 t y^{^{'}} - 6 y = 0

\int \frac{1}{3 + t} d t

y^{^{'}} = 3 y x^{2} - 3 x^{2}

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

What is the tangent of f(x)=sqrt(3x+1),(5,4) ?
The tangent of f(x)=sqrt(3x+1),(5,4) is y= 3/8 x+17/8