Mathematics, 25.01.2020 14:31 chloemandile9818

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

For brainliest: finding critical points & extreme values. #47 & #61 provide a clear expla

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Answer from: SmokeyRN

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).

Answer from: neirabrandon516

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
For brainliest: finding critical points & extreme values. #47 & #61 provide a clear expla

For brainliest: finding critical points & extreme values. #47 & #61 provide a clear expla

Answer from: Quest

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all

Answer from: Quest

(23*2)-12=x x=34

step-by-step explanation:

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For brainliest: finding critical points & extreme values. #47 & #61 provide a clear expla...