What is the equation of the tangent line of $f(x)=sqrt(x+1)$ at $x=0$ ?

Question

4982 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-13T00:36:27

$2y-x-2=0$

We have $f(0)=1$ , and

$f^'(x) = 1/(2sqrt(x+1))$

and thus

$f^'(0) = 1/2$

So, the required tangent is a straight line with slope $1/2$ , passing through $(0,1)$

It has the equation

$y-1 = 1/2(x-0) implies 2y-x-2=0$

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