What is the tangent of f(x)=sqrt(x+1),\at x=0 ?

The tangent of f(x)=sqrt(x+1),\at x=0 is y= 1/2 x+1

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+1),\at x=0

tangent of

f (x) = x + 1, at x = 0

y = \frac{1}{2} x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(0, 1) : y = \frac{1}{2} x + 1

y = \frac{1}{2} x + 1

f (x) = e^{x}

\int sin (4 x) cos (6 x) d x

\frac{d}{d x} (3 x - 4)

- 0.01549944 x^{2} + 0.287342 x

\frac{d}{d x} (\frac{10 x + 8 x ^{2}}{( 5 + 8 x ) ^{2}})

What is the tangent of f(x)=sqrt(x+1),\at x=0 ?
The tangent of f(x)=sqrt(x+1),\at x=0 is y= 1/2 x+1