Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 20

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

The guidelines of Curve Sketching
A. Domain.
We know that $f(x)$ is a root function that is defined only for positive value of $x$. Therefore, the domain is $[0,\infty)$

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

$
\begin{equation}
\begin{aligned}
0 & = 2 \sqrt{x} - x \\
\\
x &= 2x^{\frac{1}{2}} = 2^2\\
\\
x &= \sqrt{4}
\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 asymptotes

E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$

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


when $y' = 0$,

$
\begin{equation}
\begin{aligned}
0 &= \frac{1}{\sqrt{x}} -1\\
\\
\sqrt{x} &= 1\\
\\
x &= 1^2
\end{aligned}
\end{equation}
$

The critical number is $x = 1$
Hence, the intervals of increase or decrease are.

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



F. Local Maximum and Minimum Values.
Since $f'(x)$ decreases from positive to negative at $x=1$, then $f(1) = 1$ is a local maximum.

G. Concavity and Points of Inflection.

$
\begin{equation}
\begin{aligned}
\text{if } f'(x) &= \frac{1}{\sqrt{x}} - 1 = x^{\frac{-1}{2}} - 1 , \text{ then}\\
\\
f''(x) &= \frac{-1}{2}x^{\frac{-3}{2}}\\
\\
f''(x) &= \frac{-1}{2\sqrt{x^3}} \\
\\
\\
\\
\text{when } f''(x) &= 0 \\
\\
0 &= \frac{-1}{2 \sqrt{x^3}}\\
\\
f''(x) &= 0 \qquad \text{ does not exist, therefore, we don't have inflection points.}
\end{aligned}
\end{equation}
$


Thus, the concavity in the domain of $f$ is...

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


H. Sketch the Graph.