Therefore, the lines $x=1$ and $x=-1$

$(1)$

E$.f^{\text{'}}(x)=\frac{4x(x^2-1)-2x^2*2x}{(x^2-1{)}^2}=0$

Since $f^{\text{'}}(x)>0$ when $)(-1$ and $f^{\text{'}}(x)<mn\; id="EditorElement\_1186213">0</mn>$ when $)(1$ is increasing on the intervals $(-,-1)$ and and decreasing on the intervals $(0,1)$ and

F$.$ The only critical number is $x=0.$ Since $f^{\text{'}}$ changes from positive to negative at $,f()=$ is a local maximum by the First Derivative Test.

G$.f^{\text{'}\text{'}}(x)=\frac{-4(x^2-1{)}^2+4x*2(x^2-1)2x}{(x^2-1{)}^4}=$

Since the simplified numerator, $>0$ for all $x,$ we have the following.

$f^{\text{'}\text{'}}(x)>0>x^2-1>0>|x|>$

and

Thus. the curve is concave upward on the intervals $(-,-1)$ and and concave downward on $.$ It has no point of inflection since $1$ and are not in the domain of $f.$