47)
[tex]\bf f(x)=x(4-x)^3\implies \cfrac{df}{dx}=(4-x)^3+x[3(4-x)^2(-1)]
\\\\\\
\cfrac{df}{dx}=(4-x)^3-3x(4-x)^2\\\\
-----------------------------\\\\
(4-x)^3\implies 64-48x+12x^2-x^3
\\\\\\
-3x(4-x)^2\implies -48x+24x^2-3x^3
\\\\\\
(64-48x+12x^2-x^3)+(-48x+24x^2-3x^3)\\\\
\implies 64-96x+36x^2-4x^3\\\\
-----------------------------\\\\[/tex]
[tex]\bf \cfrac{df}{dx}=64-96x+36x^2-4x^3\implies 0=64-96x+36x^2-4x^3
\\\\\\
0=x^3-9x^2+24x-16\impliedby
\begin{array}{llll}
\textit{now, doing a synthetic division}\\
\textit{with x=4, gives us}
\end{array}
\\\\\\
0=(x-4)(x^2-5x+4)\implies 0=(x-4)(x-4)(x-1)
\\\\\\
x=
\begin{cases}
4\\
1
\end{cases}[/tex]
now, on 61)
[tex]\bf y=\cfrac{x}{x^2+1}\implies \cfrac{dy}{dx}=\cfrac{(x^2+1)-x\cdot 2x}{(x^2+1)^2}\impliedby \textit{quotient rule}
\\\\\\
\cfrac{dy}{dx}=\cfrac{x^2+1-2x^2}{(x^2+1)^2}\implies \cfrac{dy}{dx}=\cfrac{1-x^2}{(x^2+1)^2}[/tex]
now, you get critical points by making the derivative to 0, OR where the denominator is 0 (cusps or asymptotes), now, for this denominator, it has no critical points, since the values it gives are imaginary ones
as far as setting the derivative to 0, [tex]\bf 0=\cfrac{1-x^2}{(x^2+1)^2}\implies 0=1-x^2\implies x=\pm\sqrt{1}\implies \boxed{x=\pm1}[/tex]
and those are the only critical points...now, if you run a first-derivative test on it, pick a value on those regions, say -2, 0 and 2, you'd get values of negative, positive and negative, check the picture below
and ... surely you can see where the extrema are at, and what they are
