Question

For BRAINLIEST:Finding Critical points & Extreme values. #47 & #61 Please provide a clear explanation that is understandable in order to get BRAINLIEST

Answer 1

47)

$\bf f(x)=x(4-x)^3\implies \cfrac{df}{dx}=(4-x)^3+x[3(4-x)^2(-1)] \\\\\\ \cfrac{df}{dx}=(4-x)^3-3x(4-x)^2\\\\ -----------------------------\\\\ (4-x)^3\implies 64-48x+12x^2-x^3 \\\\\\ -3x(4-x)^2\implies -48x+24x^2-3x^3 \\\\\\ (64-48x+12x^2-x^3)+(-48x+24x^2-3x^3)\\\\ \implies 64-96x+36x^2-4x^3\\\\ -----------------------------$

$\bf \cfrac{df}{dx}=64-96x+36x^2-4x^3\implies 0=64-96x+36x^2-4x^3 \\\\\\ 0=x^3-9x^2+24x-16\impliedby \begin{array}{llll} \textit{now, doing a synthetic division}\\ \textit{with x=4, gives us} \end{array} \\\\\\ 0=(x-4)(x^2-5x+4)\implies 0=(x-4)(x-4)(x-1) \\\\\\ x= \begin{cases} 4\\ 1 \end{cases}$

now,  on 61)

$\bf y=\cfrac{x}{x^2+1}\implies \cfrac{dy}{dx}=\cfrac{(x^2+1)-x\cdot 2x}{(x^2+1)^2}\impliedby \textit{quotient rule} \\\\\\ \cfrac{dy}{dx}=\cfrac{x^2+1-2x^2}{(x^2+1)^2}\implies \cfrac{dy}{dx}=\cfrac{1-x^2}{(x^2+1)^2}$

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, $\bf 0=\cfrac{1-x^2}{(x^2+1)^2}\implies 0=1-x^2\implies x=\pm\sqrt{1}\implies \boxed{x=\pm1}$

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

Answer 2

Start with #47.  To find the critical values, you must differentiate this function.  x times (4-x)^3 is a product, so use the product rule.  The derivative comes out to f '(x) = x*3*(4-x)^2*(-1) + (4-x)^3*1 = (4-x)^2 [-3x + 4-x]
Factoring this, f '(x) = (4-x)^2 [-3x+4-x] 
Set this derivative equal to zero (0) and solve for the "critical values," which are the roots of   f '(x) = (4-x)^2 [-3x+4-x].  (4-x)^2=0 produces the "cv" x=4.
[-3x+ (4-x)] = 0 produces the "cv" x=1.   Thus, the "cv" are {4,1}.

Evaluate the given function at x: {4,1}.  For example, if x=1, f(1)=(1)(4-1)^3, or 2^3, or 8.  Thus, one of the extreme values is (1,8).