Answer
$y(1)=0$ is a local minimum, and $y(-1)=0$ is a local maximum.
no absolute extrema.
inflection points: $x=-0.92,0.69$
concave down on $(-0.92,-0.5),(0.69,1)$ and concave up on $(-1,-0.92),(-0.5,0.69)$.
See graph.
Work Step by Step
Step 1. Given the function $y=\frac{\sqrt {1-x^2}}{2x+1}=(1-x^2)^{1/2}(2x+1)^{-1}, -1\leq x \leq1, x\ne-1/2$, we have $y'=\frac{1}{2}(1-x^2)^{-1/2}(2x+1)^{-1}(-2x)-2(1-x^2)^{1/2}(2x+1)^{-2}=\frac{-x}{\sqrt {1-x^2}(2x+1)}-\frac{2\sqrt {1-x^2}}{(2x+1)^2}=\frac{-x(2x+1)-2(1-x^2)}{\sqrt {1-x^2}(2x+1)^2}=\frac{-x-2}{\sqrt {1-x^2}(2x+1)^2}=-(x+2)(1-x^2)^{-1/2}(2x+1)^{-2}$ and $y''=-(1-x^2)^{-1/2}(2x+1)^{-2}+\frac{1}{2}(x+2)(1-x^2)^{-3/2}(2x+1)^{-2}(-2x)+2(x+2)(1-x^2)^{-1/2}(2x+1)^{-3}(2)=\frac{-1}{\sqrt {1-x^2}(2x+1)^2}+\frac{-x^2-2x}{\sqrt {(1-x^2)^3}(2x+1)^2}+\frac{4x+8}{\sqrt {1-x^2}(2x+1)^3}=\frac{-(1-x^2)(2x+1)-(x^2+2x(2x+1))+(4x+8)(1-x^2)}{\sqrt {(1-x^2)^3}(2x+1)^3}=\frac{-4x^3-12x^2+7}{\sqrt {(1-x^2)^3}(2x+1)^3}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. The possible critical points are $x=-2, -1,-1/2, 1$ and we need to exclude $x=-2,-1/2$ as they are not in the domain. We have $y(-1)=y(1)=0$
Step 3. Check sign changes of $y'$ across the critical point $(0,0)$, $(-1)..(-)..(-1/2)..(-)..(1)$. We can identify $y(1)=0$ as a local minimum, and $y(-1)=0$ as a local maximum. The function has no absolute extrema due to the asymptote at $x=-1/2$.
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We have $4x^3+12x^2-7=0$ which gives $x=-0.92,0.69$ in $[-1,1]$ (solve the equation graphically as necessary).
Step 5. Examine the sign of $y''$ on different intervals. We have $(-1)..(+)..(-0.92)..(-)..(-1/2)..(+)..(0.69)..(-)..(1)$, The function is concave down on $(-0.92,-0.5),(0.69,1)$ and concave up on $(-1,-0.92),(-0.5,0.69)$.
Step 6. See graph.
