Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 322: 26

A. The domain is $[-\sqrt{2}, \sqrt{2}]$ B. The y-intercept is $0$ The x-intercepts are $-\sqrt{2}, 0, \sqrt{2}$ C. The function is an odd function. D. There are no asymptotes. E. The graph is decreasing on the intervals $(-\sqrt{2},-1)\cup (1,\sqrt{2})$ The graph is increasing on the interval $(-1,1)$ F. The local minimum is $(-1,-1)$ The local minimum is $(1,1)$ G. The graph is concave up on the interval $(-\sqrt{2},0)$ The graph is concave down on the interval $(0, \sqrt{2})$ The point of inflection is $(0,0)$ H. We can see a sketch of the curve below.

$y = x\sqrt{2-x^2}$ A. The function is defined for real numbers such that $(2-x^2) \geq 0$: $x^2 \leq 2$ $-\sqrt{2} \leq x \leq \sqrt{2}$ The domain is $[-\sqrt{2}, \sqrt{2}]$ B. When $x = 0$, then $y = (0)\sqrt{2-0^2} = 0$ The y-intercept is $0$ When $y = 0$: $x\sqrt{2-x^2} = 0$ $x = -\sqrt{2}, 0, \sqrt{2}$ The x-intercepts are $-\sqrt{2}, 0, \sqrt{2}$ C. $(-x)\sqrt{2-(-x)^2} = -x\sqrt{2-x^2}$ The function is an odd function. D. There are no asymptotes. E. We can find values of $x$ such that $y' = 0$: $y' =\sqrt{2-x^2}+(x)~(\frac{-x}{\sqrt{2-x^2}})$ $y' =\frac{2-2x^2}{\sqrt{2-x^2}} = 0$ $2-2x^2 = 0$ $x = -1, 1$ When $-\sqrt{2} \lt x \lt -1$ or $1 \lt x \lt \sqrt{2}$, then $y' \lt 0$ The graph is decreasing on the intervals $(-\sqrt{2},-1)\cup (1,\sqrt{2})$ When $-1 \lt x \lt 1$, then $y' \gt 0$ The graph is increasing on the interval $(-1,1)$ F. When $x = -1$, then $y = (-1)\sqrt{2-(-1)^2} = -1$ The local minimum is $(-1,-1)$ When $x = 1$, then $y = (1)\sqrt{2-(1)^2} = 1$ The local minimum is $(1,1)$ G. We can find the values of $x$ such that $y'' = 0$: $y'' =\frac{(-4x)(\sqrt{2-x^2})-(2-2x^2)(\frac{-x}{\sqrt{2-x^2}})}{2-x^2}$ $y'' =\frac{(-4x)(2-x^2)-(2-2x^2)(-x)}{(2-x^2)^{3/2}}$ $y'' =\frac{-8x+4x^3+2x-2x^3}{(2-x^2)^{3/2}}$ $y'' =\frac{2x^3-6x}{(2-x^2)^{3/2}} = 0$ $2x^3-6x = 0$ $2x(x^2-3) = 0$ $x = 0$ Note that $x = \pm \sqrt{3}$ are not included in the domain. When $-\sqrt{2} \lt x \lt 0$, then $y'' \gt 0$ The graph is concave up on the interval $(-\sqrt{2},0)$ When $0 \lt x \lt \sqrt{2}$, then $y'' \lt 0$ The graph is concave down on the interval $(0, \sqrt{2})$ When $x = 0$, then $y = (0)\sqrt{2-0^2} = 0$ The point of inflection is $(0,0)$ H. We can see a sketch of the curve below.