For brainliest: finding critical points & extreme values. #47 & #61 provide a clear explanation that is understandable in order to get brainliest

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).  

47)





now,  on 61)



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, 

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

The third box is a function

The answer would be 484