Answer
- no extrema.
- inflection point is $(0,0)$
- concave up on $(-\infty,0)$ and concave down on $(0,\infty)$.
See graph.
Work Step by Step
Step 1. Given the function $y=\frac{x}{\sqrt {x^2+1}}=x(x^2+1)^{-1/2}$, we have $y'=(x^2+1)^{-1/2}-\frac{1}{2}x(x^2+1)^{-3/2}(2x)=(x^2+1)^{-1/2}-x^2(x^2+1)^{-3/2}=\frac{x^2+1-x^2}{\sqrt {(x^2+1)^3}}=\frac{1}{\sqrt {(x^2+1)^3}}$ and $y''=-\frac{1}{2}(x^2+1)^{-3/2}(2x)-2x(x^2+1)^{-3/2}+\frac{3}{2}x^2(x^2+1)^{-5/2}(2x)=-x(x^2+1)^{-3/2}-2x(x^2+1)^{-3/2}+3x^3(x^2+1)^{-5/2}=-3x(x^2+1)^{-3/2}+3x^3(x^2+1)^{-5/2}=\frac{3x^3-3x(x^2+1)}{\sqrt {(x^2+1)^5}}=\frac{-3x}{\sqrt {(x^2+1)^5}}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. As $y'\ne0$ and there are no endpoints or undefined points, the function has no extrema.
Step 3. The inflection points can be found when $y''=0$ or when it does not exist. We have $x=0$ and the possible inflection point is $(0,0)$
Step 4. To identify the intervals on which the functions are concave up and concave down, we need to examine the sign of $y''$ on different intervals. We have $..(+)..(0)..(-)..$,
Step 5. The function is concave up on $(-\infty,0)$ and concave down on $(0,\infty)$.
Step 6. See graph.
