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

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

Answer from: Quest

days of the week

step-by-step explanation:

javier earned more money then kyle, so that won't always be a constant term, but the days of the week do not change so that term is always constant.

edit: i meant

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