Question

Find all roots, local maxima and minima, and inflection points of each function f . In addition, determine whether any local extrema are also global extrema on the domain of f .

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

Short answer

function has no root

function has no minima point

function has no maxima point

function has no inflection point

Step-by-step solution

Step1. Given Information

Consider a function

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

For finding roots of the function.

$\begin{array}{r}f\left(x\right)=0\\ \frac{1}{1+\sqrt{x}}=0\end{array}$

The function has no root.

Step2. Critical point

The first derivative of the function is given below.

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=\frac{d}{dx}\left(\frac{1}{1+\sqrt{x}}\right)\\ =-\frac{1}{(1+\sqrt{x}{)}^2}\end{array}$

$\text{This derivative is always defined and continuous, so the critical points of}f\text{are just the places}$

where $f\text{'}\left(x\right)=0;$that is

$\begin{array}{c}-\frac{1}{(1+\sqrt{x}{)}^2}=0\\ \frac{1}{(1+\sqrt{x}{)}^2}=0\end{array}$

The function has no critical point.

Step3. Inflection point

$\begin{array}{r}{f}^{\prime \prime }\left(x\right)=0\\ \frac{d}{dx}\left[-\frac{1}{(1+\sqrt{x}{)}^2}\right]=0\\ \frac{d}{dx}\frac{1}{(1+\sqrt{x}{)}^2}=0\\ -2(1+\sqrt{x}{)}^{-2-1}\frac{1}{2\sqrt{x}}=0\\ \frac{1}{\sqrt{x}(1+\sqrt{x}{)}^3}=0\end{array}$

Therefore, the function has no inflection point.

Step4. Answer

Therefore, the function does not have a root; and the function has neither minimum nor maximum. Therefore, the function has no inflection point. It has no global extrema.