June 17, 20XX
The forces of good clashed with the mechanical might of Metallix and his legion of robotic henchmen. The air was filled with tension as Mario, Sonic, Luigi, Shadow, and Yoshi prepared to defend the Mushroom Kingdom from this formidable threat.
As the battle commenced, Metallix, the cunning and ruthless metallic version of Sonic, proved to be a formidable adversary. With his advanced speed and agility, he effortlessly weaved through the chaos, engaging in fierce combat with our heroes. But that wasn't all - he invoked the aid of another robotic henchman, Axem Blue, to intensify the fight.
Axem Blue, a robotic warrior known for his incredible strength and precision, joined the battle with an intimidating presence. His metallic frame glimmered under the sunlight, and his axe-like arms demonstrated his mastery of close-quarters combat. It seemed as though Metallix had learned from previous encounters and was determined to test our heroes to their limits.
However, our heroes were not alone in this fight. As the situation grew dire, Robo Mario, a mechanized version of our courageous plumber, emerged to assist in turning the tide. Coordinating their efforts, Metallix and Mecha Mario engaged together against them in a ferocious showdown, their metallic clashes resonating throughout the raceways.
Robo Mario, as a loyal companion of our heroes, realized the gravity of the situation and promptly coordinated its attacks with Mario, Sonic, Luigi, Shadow, and Yoshi. With their combined strength, they formed an unbreakable bond, ready to face any challenge that came their way.
The Battle of Mushroom Raceways raged on, with sparks flying, and the ground trembling beneath the weight of the combatants. Each strike, each maneuver, was a testament to the unyielding determination of both sides. Metallix and his henchmen fought relentlessly, their robotic precision matching the sheer willpower and teamwork of our heroes.
In the end, it was the resilience and unity of our heroes that prevailed. They overcame the mechanical might of Metallix and his allies, gradually gaining the upper hand. Together, they unleashed a barrage of coordinated attacks, exploiting the weaknesses of their opponents, until the metallic army was forced to retreat.
Math Notes: (filled with academic musings)
Finding the tangent
y=-x^3 + 8x^2 - 20x + 14 at (2,-2)
dy/dx or slope (m) = -3x^2 + 16x - 20 at (2,-2)
m= -3(2)^2 + 16(2) - 20
m=0 or y'=0
y=mx+b
y=(0)x+b
y=(0)2-2
b=-2
Answer: y=-2
...
y= 6 / x^2+2 at (3, 6/11)
dy/dx = (x^2 + 2) d/dx(6) - (6) d/dx(x^2+2) / (x^2 + 2)^2
dy/dx = (x^2 + 2) (0) - (6) (2x+0) / (x^2 + 2)^2
dy/dx = (3^2 + 2) (0) - (6)(2 times 3+0) / (3^2 + 2)^2
dy/dx: m=-36/121
y=mx+b
y=(-36/121)x+b
(3, 6/11)
6/11=(-36/121)(3)+b
174/121=b
y=-36/121x + 174/121
...
y=-2sin(x) at (0,0)
dy/dx=-2cos(x)
m=-2cos(0)
m=0
-2cos(0)
=-2(1)
m=-2
y=(-2)x+b
y=(-2)(0)+b
0=(-2)(0)+0
b=0
y=(-2)x+0
...
y=(2x-4)^1/3 at (-2,-2)
dy/dx=1/3 (2x-4)^1/3-1
f'(x)=f(g(x)) Chain Rule
1/3 (2x-4)^1/3 - 1 (2)
dy/dx=1/3 (2x-4)^-2/3 (2)
(-2,-2)
x=-2
dy/dx=1/3 (2(-2)-4)^-2/3 (2)
dy/dx=1/6
m=1/6
y=(1/6)x+b
(-2)=(1/6)(-2)+b
b=-7/3
y=(1/6)x - 7/3
...
y= -3sin(x) at (0,0)
dy/dx=-3sin(x)
m=-3cos(0)
m=0
cos(0)=1
So: -3cos(0)
=-3(1)
m=-3
y=(-3)x+b
y=(-3)(0)+b
b=0
y=(-3)x+0
...
y=(2x-4)^1/4 at (-2,-2)
y=(3x-5)^1/4 at (-3,-3)
dy/dx=1/4(3x-5)^1/4-1
f'(x)=f(g(x)) chain rule
dy/dx=1/4(3x-5)^1/4 - 1 (3)
(-3,-3)
dy/dx=1/4(3(-3)-5)^1/4 - 1 (3)
dy/dx=1/4(3(-3)-5)^-5/4 (3)
Domain Error
...
y=(3x-5)^1/2 at (-3,-3)
dy/dx = 1/2(3x-5)^1/2-1
f'(x)=f(g(x)) chain rule
dy/dx=1/2(3x-5)^-1/2 (3)
dy/dx=1/2(3(-3)-5)^-3/2(3)
Domain Error
...
Mean Value Theorm
Continious [a,b] vs (a,b)
f'(c)=f(b)-f(a) / b-a
mtan=msec
m=y2-y1 / x2-x1
f(x)=y
f(b)=y2
f(a)=y1
...
f'(c)= f(b)-f(a) / b-a
f(x)=2x^3 + x^2 + 7x - 1 on [1,6]
f(a)=2(1)^3 + (1)^2 + 7(1) - 1
f(a)=9
f(b)=2(6)^3 + (6)^2 + 7(6) - 1
f(b)=509
509-9 / 6-1 = 100
f'(c)=100
Derivative
dy/dx = f'(x) = 6x^2 + 2x + 7
100=6x^2 + 2x + 7
0=6x^2 + 2x - 93
-b+/- square root b^2 - 4(a)(c) / 2(a)
-2+/- square root 2^2 - 4(6)(-93) / 2(6)
=3.77... and -4.11...
x=3.77...
Plug in the x to the f(x)'s x
=(3.77... , 147.15...)
...
f(z)=4z^3 - 8z^2 + 7z - 2 on [2,5]
f(a)=4(2)^3 - 8(2)^2 + 7(2)-2
f(a)=12
f(b)=4(5)^3 - 8(5)^2 + 7(5) - 2
f(b)=333
333-12 / 5-2 = 321/3 = 107
f'(c)=107
dy/dx=f'(z)=12z^2 - 16z + 7
107=12z^2 - 16z + 7
0=12z^2 -16z - 100
Since difficult to factor: -b+/- sqaure root b^2 - 4(a)(c) / 2(a)
-(-16)+/- square root (-16)^2 - 4(12)(-100) / 2(12)
x= 3.63...
x=3.63...
y=4(3.63...)^3 - 8(3.63...)^2 + 7(3.63...) - 2
y=109.26...
(3.63... , 109.26...)
...
1) Find the derivative set = 0. Solve for x
2) Create a number line and plug in points close to the extremas to determined if the extremas are relative with minimums and maximums.
3) Plug in the endpoints as well as the relative extremas.
4) Determine the absolute extrema by carrying y-values
f(x)=x^2 - 4x + 9; [1,4]
x=2
f(1)=1^2 - 4(1) + 9
1-4+9=6
f(2)=2^2 - 4(2) + 9
4-8+9=5
f(4)=4^2 - 4(4) + 9
16-16+9=9
f(x)=x^2-4x+9
f'(x)=2x-4=0
0=2(x-2)
x=2
Number line: Before 2 has negative range, after 2 has positive range
(1,6)
(2,5): absolute min
(4,9): absolute max
...
y=-x^3 + x^2 - 3; [-1,1]
y'=-3x^2 + 2x
0=-3x^2 + 2x
x(-3x+2)
x=0, x=2/3
f(-1/3)=-3(-1/3)^2 + 2(-1/3)
f(-1/3)=-1
f(1)=-1
f(1/3)= -3(1/3)^2 + 2(1/3) = 1/3
Max: (-1,-1)
(2/3, -2.85...)
Min: (0,3), (1-3)