What is the tangent of f(x)=sqrt(2x+1),\at x=3 ?

The tangent of f(x)=sqrt(2x+1),\at x=3 is y= 1/(sqrt(7))x+sqrt(7)-3/(sqrt(7))

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(2x+1),\at x=3

tangent of

f (x) = 2 x + 1, at x = 3

y = \frac{1}{7} x + 7 - \frac{3}{7}

Solution steps

Find the tangent point:

(3, 7)

Find the slope of

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

Compute the slope of

f (x) = 2 x + 1

x = 3 : m = \frac{1}{7}

Find the line with slope m=

\frac{1}{7}

and passing through

(3, 7) : y = \frac{1}{7} x + 7 - \frac{3}{7}

y = \frac{1}{7} x + 7 - \frac{3}{7}

g (t) = tan (cos (2 t))

(- 1, - 1), (1, 5)

\frac{d}{d x} ((3 x - 1)^{4} (2 x + 1)^{- 3})

x y^{^{'}} - y = x^{2} + 5 x - 1

f (x) = 3 x^{5} - 2 x^{4}, at x = 4

What is the tangent of f(x)=sqrt(2x+1),\at x=3 ?
The tangent of f(x)=sqrt(2x+1),\at x=3 is y= 1/(sqrt(7))x+sqrt(7)-3/(sqrt(7))