How do you find the equation of the line tangent to $f (x) = (\sqrt{x} + 1)$ $f(x)= (sqrtx+1)$ , at (0,1)?

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Answer 1 · 2016-01-26T21:31:45

Find the derivative at the point (= the slope of the graph's tangent). Then find the equation with the slope and point that you have.

First: the slope=the derivative= $\frac{d}{d x} [f (x)]$ $d/dx[f(x)]$
$\frac{d}{d x} [\sqrt{x} + 1] = \frac{d}{d x} [x^{\frac{1}{2}} + 1]$ $d/dx[sqrt(x)+1]=d/dx[x^(1/2)+1]$

$= \frac{1}{2} \cdot x^{- \frac{1}{2}}$ $=1/2*x^(-1/2)$ $= \frac{1}{2} \cdot \frac{1}{\sqrt{x}}$ $=1/2*1/sqrt(x)$

$= \frac{1}{2 \sqrt{x}}$ $=1/(2sqrt(x))$

The slope of the line tangent to f(x) at $(0, 1) =$ $(0,1)=$
$\frac{1}{2 \sqrt{0}} = \frac{1}{0} \Rightarrow$ $1/(2sqrt(0))=1/0=>$ the line is parallel to the $y$ $y$ -axis
Second: the equation of the tangent is
$x = a$ $x=a$ where $a$ $a$ is constant
again it passes through (0,1)
so equation should be $x = 0$ $x=0$ i.e. $y$ $y$ axis

How do you find the equation of the line tangent to f(x)=(√x+1)f(x)= (sqrtx+1), at (0,1)?