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How Do You Differentiate Using Relative Extrema I Have Been Stuck On This One Problem For A While

Use the first derivative test to find relative extrema of the function: g(x) = (x^3-8x)/4?

Differentiate the function g(x)...

g'(x) = ¼ * (3x² - 8)

Let g'(x) = 0 [since this is where relative extremum occurs], so..

0 = 3x² - 8
3x² = 8
x² = 8/3
x = ±√(8/3)

Using the first derivative test, we determine where g'(x) > 0 [positive] and g'(x) < 0 [negative].

Original function: http://www.wolframalpha.com/input/?i=%28x^3-8x%29%2F4
First Derivative Test: http://www.wolframalpha.com/input/?i=differentiate+%28x^3-8x%29%2F4

g'(x) > 0 [increasing] where x < -√(8/3) and x > √(8/3)
g'(x) < 0 [decreasing] where -√(8/3) < x < √(8/3)

We now have max at x = -√(8/3) and min at x = √(8/3) [Noting that g'(x) = 0 shows stopping increase and decrease of g(x). IT would be better to study the pattern between the graphs of g'(x) and g(x)]

I hope this helps!

Relative Extrema and Second-Derivative Test?

f(x) = x^3 - 5x^2 + 7x --> f'(x) = 3x^2 - 10x + 7
The critical values at where f'(x) = 0 =(3x - 7)(x - 1) --> x = 7/3 or 1
f"(x) = 6x - 10 --> f"(7/3) = 6(7/3) - 10 = 4 > 0 --> minimum
--> f"(1) = 6(1) - 10 = -4 < 0 --> maximum


f(x) = x^(2/3) - 3 --> (2/3)x^(-1/3) f'(x) is undefined at x = 0 (critical value)
f"(x) = (2/3)(-1/3)x^(-4/3) which is undefined at x = 0 so the second derivative test cannot be used

the first derivative is negative if x < 0 --> decreasing function
the first derivative is positive if x > 0 --> increasing function
This implies that there is a minimum value at x = 0

Finding relative extrema when given the derivative ?

Set the derivative equal to zero and solve for x:
0 = sin(x² + 1)
kπ = x² + 1
kπ - 1 = x²
±√(kπ - 1) = x

K is an integer, so plug in values for k and solve for x-values in between 2 and 4. When k = 2, 3, 4, and 5 it works out, thus:
x = √(2π - 1), √(3π - 1), √(4π - 1), and √(5π - 1) are critical points

Find any relative extrema of the function? With Hyperbolic Functions ?

Find f'(x) then set it = 0.

f(x) = sin * sinh - cos * cosh

f'(x) = (cos * sinh + sin * cosh) - (-sin * cosh - cos * sinh) = 2 * (cos * sinh + sin * cosh).

When x = 0, both sin(x) and sinh(x) = 0, so we know that x=0 is a local extreme. Just look for others in the interval in question.

Do linear functions and constants have a relative maxima/minima?

not linear functions but all points of a constant are both relative and absolute max/min

functions whose derivatives change sign twice { on the domain } will have both relative max/min

Ax^3 + bx^2 + c; find a,b,c given local extremum and inflection pt?

the second derivative of f(x) = ax^3 +bx^2 +c is f '' (x) = 6ax+2b and 1 is a root (zero) of f ''
then 6a+2b = 0
first derivative of f (x) is 3ax^2+2bx and 2 is a root of f ' (x)
then 12a+4b = 0
and
f(2) = 11
then 8a+4b+c = 11
f(1) = 5
a+b+c = 5

Now solve
a+b+c = 5
8a+4b+c = 11
3a+b = 0

7a+3b = 6
3a+b = 0

2a = -6
then a = -3
b = 9
c = -1
f(x) = -3x^3 +9x^2 -1

Plate stuck inside a bowl?

Something similar happened to us. A plastic character plate got stuck in a pot and finally we found something that worked to get it out. A family member took it outside and turned on the hose with an attached sprayer. He sprayed into the crack and the high power eventually went to the other side shooting up and loosened the plate. I didn't want to break the plate (collectible from a very long time ago) and didn't want to damage the pot. Try it and see if it works. Sticking things into it may or may not work. A lot of times I get cups stuck in the sink drain that the garbage disposal is in and have a hard time getting it cups out to continue washing dishes. belong in a disposal in there. Good luck.

The idea of the hose is to make sure that the higher pressure is going inside the crack to help force the pressure going up on the opposite side of the plate. We watched the plate slowly rise up since we couldn't grab it as there wasn't any room for us to get something in there and when it came to the top it floated up enough above the water for us to grab. Also the water went in easily through the crack unlike an object would. I don't know if hot water will work cause normally when a jar lid is hard to get off the lid will expand after being held in water but I don't think a plate would change shape without breaking. I did leave the plate in the pot for about a week till we figured out a way to get it out while we used another pot.

What is the difference between a wormhole and a black hole?

BlackholesIt is a region of space-time so deformed by its mass that it exerts a gravitational pull that not even light can escape. The point at which the gravitational pull becomes so great as to make escape impossible is called the event horizonAt the center of the blackhole lies a region of space-time with infinite curvature.Blackholes are formed when the core of a supermassive star collapses in on itselfWormholesTheir official nomenclature is "Einstein-Rosen bridge" and are a hypothetical tunnel that would connect two points in space-time.There are valid solution for equations in the theory of relativity that contain wormholes, but unlike black holes, there hasn't been any observational evidence for wormholes so far.In theory, a stable wormhole could be used as a way to travel across vast expanses of space.

Who suffers more due to tension between wife and mother-in-law, the wife or the husband?

I would have to say the husband and mother-in-law suffer the most because the husband is forced to listen to his wife insult his mother. Even kind insults are hurtful. He should and does put his wife first but his mother is his roots and he can’t cut her off without cutting his own beginnings and life force. When a couple marries it doesn’t mean they shut everyone else out. They make their own plans and decisions but that does not include being mean or and to another person especially if it is the person’s close relative. Mothers-in-law can be annoying like a child but one must learn how to deal with this as they learn how to evade questions from a child. Hurting another is not necessary. In the end we most likely will be in both roles. We can’t imagine it when we are young in love and ready to be independent and free. It is not required that you constantly give in but that you always respect and nicely say no when you have to but also bend when you can. Husbands will be less anxious if both women in his life are happy. He loves his wife first but he loves his mother whether he cuts all ties with her for his wife’s sake, or not. I wrote about mothers/daughters-in-law in my book THE PRINCESS AND THE QUEEN. I share over twenty stories about real women. There are questions reflections and quotes about all kinds of problems that arise in this relationship. One man after ten years of not seeing or speaking with his mother, got in contact with her. He was sorry and ended up indulging his mother in many ways as his form of repentance. It is cruel to shut her out of your lives but not cruel to disagree with her interference and do your own thing. We need to treat all people with respect. You can’t go wrong if you live this way. We all have feelings and opinions. Our backgrounds are different and they color our opinions. Make an attempt to compromise. Peace always makes one calmer whereas fighting and tension brings upheaval and more fighting. I choose serenity. God Bless and lots of luck