# How do you graph y =(x^(1/2) +1)/(x+1)?

Sep 18, 2015

Analyse this function to find the domain, some points through which it goes, maximum and asymptotic behaviour.

#### Explanation:

Let $f \left(x\right) = \frac{{x}^{\frac{1}{2}} + 1}{x + 1}$

${x}^{\frac{1}{2}}$ only takes Real values for $x \ge 0$.

Hence the domain of $f \left(x\right)$ is $\left[0 , \infty\right)$

$f \left(0\right) = f \left(1\right) = 1$

$f ' \left(x\right) = \frac{- {x}^{\frac{1}{2}} + {x}^{- \frac{1}{2}} - 2}{2 {\left(x + 1\right)}^{2}}$

Let $t = {x}^{\frac{1}{2}}$

Then $f ' \left(x\right) = 0$ when $- t + \frac{1}{t} - 2 = 0$

Multiply this equation through by $- t$ to get:

${t}^{2} + 2 t - 1 = 0$

This has roots:

$t = \frac{- 2 \pm \sqrt{{2}^{2} - \left(4 \times 1 \times - 1\right)}}{2 \cdot 1} = - 1 \pm \sqrt{2}$

Since $t = \sqrt{x} \ge 0$, $\sqrt{x} = - 1 + \sqrt{2}$.

So $x = {\left(\sqrt{2} - 1\right)}^{2} = 3 - 2 \sqrt{2} \approx 0.17157$

When $x = 3 - 2 \sqrt{2}$:

$f \left(x\right) = \frac{\sqrt{x} + 1}{x + 1} = \frac{\sqrt{2}}{4 - 2 \sqrt{2}} = \frac{\sqrt{2} \left(2 + \sqrt{2}\right)}{2 \left(2 - \sqrt{2}\right) \left(2 + \sqrt{2}\right)}$

$= \frac{\sqrt{2} + 1}{4 - 2} = \frac{\sqrt{2} + 1}{2} \approx 1.2071$

As $x \to \infty$ $f \left(x\right) \to 0$ since the denominator has higher degree than the numerator, but quite slowly due to the ${x}^{\frac{1}{2}}$ degree.

Putting this together, we have a curve only defined for $x \ge 0$, that goes through $\left(0 , 1\right)$ and $\left(1 , 1\right)$, is asymptotic to $y = 0$ and has a maximum at $\left(3 - 2 \sqrt{2} , \frac{\sqrt{2} + 1}{2}\right) \approx \left(0.17157 , 1.2071\right)$

graph{(x^(1/2)+1)/(x+1) [-0.552, 4.448, -0.78, 1.72]}