## Calculus 10th Edition

Over the specified interval, the function has an absolute maximum equal to $1$ and an absolute minimum equal to $-1.$
Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=2x; u'(x)=2$ $v(x)=x^2+1; v'(x)=2x$ $f'(x)=\dfrac{2(x^2+1)-(2x)(2x)}{(x^2+1)^2}=\dfrac{2(1-x)(1+x)}{(x^2+1)^2}.$ $f'(x)=0\to x=1$ or $x=-1.$ Along with the interval's boundaries, both $x=1$ and $x=-1$ are possible candidates for absolute extrema. $f(-2)=-\dfrac{4}{5}.$ $f(-1)=-1.$ $f(1)=1.$ $f(2)=\dfrac{4}{5}.$ Over the specified interval, the function has an absolute maximum equal to $1$ and an absolute minimum equal to $-1.$