Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 22

Use the guidelines of curve sketching to sketch the curve. $\displaystyle y = \sqrt{x^2+x} -x $

The guidelines of Curve Sketching
A. Domain.
We know that $f(x)$ contains square root function that is defined only for positive values of $x$. Hence, $x^2 + x \geq 0 \Longrightarrow x(x+1) \Longrightarrow x \geq 0$ or $x \leq -1$. Therefore, the domain is $(-\infty, -1]\bigcup[0, \infty)$

B. Intercepts.
Solving for $y$-intercept, when $x=0$,
$y = \sqrt{0^2 + 0} - 0 = 0$
Solving for $x$-intercept, when $y=0$

$
\begin{equation}
\begin{aligned}
0 & = \sqrt{x^2 + x } - x\\
\\
\sqrt{x^2+x} &= x\\
\\
x^2 + x & = x^2\\
\\
x &= 0
\end{aligned}
\end{equation}
$



C. Symmetry.
The function is not symmetric to either $y$-axis or origin by using symmetry test.


D. Asymptotes.
The function has no vertical asymptotes
For the horizontal asymptotes

$
\begin{equation}
\begin{aligned}
\lim_{x \to \pm\infty} \left(\sqrt{x^2+x} - x \right) &= \lim_{x \to \pm\infty} \left( \sqrt{x^2+x} - x \right) \cdot \left( \frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x} \right) && \Longleftarrow \text{(By multiplying the conjugate)}\\
\\
&= \lim_{x \to \pm\infty}\frac{x^2 + x - x^2}{\sqrt{x^2 + x} +x }\\
\\
&= \lim_{x \to \pm\infty} \frac{x}{\sqrt{x^2+x}+x} && \Longleftarrow \left(\text{by dividing to } \frac{1}{x}\right)\\
\\
&= \lim_{x \to \pm\infty}\frac{1}{\sqrt{1 + \frac{1}{x} }} + 1\\
\\
&= \frac{1}{2}
\end{aligned}
\end{equation}
$

Therefore, the horizontal asymptote is $\displaystyle y = \frac{1}{2}$

E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$, By using Chain Rule,

$
\begin{equation}
\begin{aligned}
y' &= \frac{1}{2} (x^2 + x)^{\frac{-1}{2}} (1x+1) -1 \\
\\
y' &= \frac{2x+1}{2\sqrt{x^2 + x}} - 1
\end{aligned}
\end{equation}
$

When $y' = 0$,
$\displaystyle 0 = \frac{2x+1}{2(x^2+x)^{\frac{1}{2}}} -1$
We don't have critical numbers.

So the intervals of increase or decrease are...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f\\
\hline\\
x < -1 & - & \text{decreasing on } (-\infty, -1)\\
\hline\\
x > 0 & + & \text{increasing on } (0, \infty)\\
\hline
\end{array}
$



F. Local Maximum and Minimum Values.
We have no local maximum and minimum.

G. Concavity and Points of Inflection.

$
\begin{equation}
\begin{aligned}
\text{if } f'(x) &= \frac{2x+1}{2\sqrt{x^2 + x }} - 1 \qquad \text{, then by using Quotient Rule and Chain Rule,}\\
\\
f''(x) &= \frac{(2\sqrt{x^2+x})(2) - (2x+1)\left(2 \left( \frac{2x+1}{2\sqrt{x^2+x}}\right) \right) }{(2\sqrt{x^2+x})^2}\\
\\
f''(x) &= \frac{4\sqrt{x^2 + x} - \frac{(2x+1)^2}{\sqrt{x^2+x}} }{(2\sqrt{x^2+x})^2} = \frac{\frac{4x^2+4x-4x^2-4x-1}{\sqrt{x^2+x}}}{4(x^2+x)}\\
\\
f''(x) &= \frac{-1}{4(x^2+)^{\frac{3}{2}}}
\end{aligned}
\end{equation}
$


when $f''(x) = 0$
$\displaystyle 0 = \frac{-1}{4(x^2+x)^{\frac{3}{2}}}$
$f''(x) = 0$ does not exist. Therefore, we don't have inflection point.
Thus, the concavity in the domain of $f$ is...


$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity}\\
\hline\\
x < -1 & - & \text{Downward}\\
\hline\\
x > 0 & - & \text{Downward}\\
\hline
\end{array}
$


H. Sketch the Graph.