Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 24

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

The guidelines of Curve Sketching
A. Domain.
We know that $f(x)$ contains square root function that is defined only for positive values. Hence,

$
\begin{equation}
\begin{aligned}
2 - x^2 &\leq 0 \\
x^2 &\geq 2
\end{aligned}
\end{equation}
$

We have,
$-\sqrt{2} \leq x \leq \sqrt{2}$
Therefore, the domain is $[-\sqrt{2},\sqrt{2}]$


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

C. Symmetry.
Since $f(-x) = -f(x)$, the function is symmetric to origin.

D. Asymptotes.
The function has no asymptotes.

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

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

when $f'(x) = 0$

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

The critical numbers are $x = 1 $ and $x = -1$

Hence, the intervals of increase and decrease are

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f\\
\hline\\
-\sqrt{2} < x < -1 & - & \text{decreasing on } (-\sqrt{2}, -1)\\
\hline\\
-1 < x < 1 & + & \text{increasing on } (-1,1)\\
\hline\\
1 < x < \sqrt{2} & - & \text{decreasing on } (1,\sqrt{2})\\
\hline
\end{array}
$


F. Local Maximum and Minimum Values.
since $f'(x)$ changes from negative to positive at $x = -1$, $f(-1) = -1$ is a local minimum. On the other hand, since $f'(x)$ changes from positive to negative at $x =1,$ $f(1) = 1$ is a local maximum.

G. Concavity and Points of Inflection.

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


when $f''(x) = 0$,
$\displaystyle 0 = \frac{-4x}{\sqrt{2-x^2}} + x(2-2x^2)$
we have $x= 0 $ as inflection point.
Thus, the concavity can be determined by dividing the interval to...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity}\\
\hline\\
-\sqrt{2} < x < 0 & - & \text{Downward}\\
\hline\\
0 < x < \sqrt{2} & + & \text{Upward}\\
\hline
\end{array}
$


H. Sketch the Graph.