June 15, 20XX
During the devastating Battle of Mushroom Raceways, the events that unfolded were nothing short of extraordinary.
As the sun rose on Mushroom Kingdom, the tranquility of the day was shattered by the thunderous roars of an impending battle. Another Koopa Troop wave was dispatched, under orders by the malevolent King Bowser. Chaos and destruction reigned as Goombas, Koopas, and other vile creatures within the army wreaked havoc, threatening the lives of innocent bystanders.
In the midst of the pandemonium, a figure clad in a vibrant blue cape emerged from the shadows. It was Geno, a mysterious warrior from Star Road, who had journeyed to the Mushroom Kingdom to restore peace and harmony. With unwavering determination and a glint of hope in his eyes, he stepped forward to confront the rampaging horde.
Geno possessed incredible powers bestowed upon him by the stars themselves were then unleashed. He summoned his immense strength, launching magical projectiles towards the invading forces, creating dazzling displays of lights that momentarily stunned his foes. With his agility and quick reflexes, he swiftly dodged enemy attacks, moving like a streak of lightning through the mayhem.
As chaos threatened to consume Mushroom Raceways, Geno shifted his focus to the civilians caught in the crossfire. His priority was to safeguard innocent lives, and he began leading them to safety, guiding them with a calm yet authoritative voice. He reassured them, instilling a sense of hope amidst the overwhelming fear and uncertainty.
In the face of overwhelming odds, Geno's determination remained unyielding. His actions inspired others to rise up and fight against the invaders, igniting a spark of courage within even the most timid of souls. Together, they formed a formidable resistance, driving back the enemy forces and reclaiming sections of the racetrack.
But Geno's role wasn't confined to leading the charge and defending the innocent. He went above and beyond, performing acts of heroism that would be forever etched in the memories of those who witnessed them. There were countless instances where Geno intercepted dangerous attacks, putting his own life on the line to shield the defenseless. He skillfully redirected fireballs and repelled enemy strikes, ensuring the safety of those under his protection.
Hours turned into an eternity as the battle raged on, but Geno's indomitable spirit never wavered. With every passing minute, he grew more resolute and relentless, pushing back the Koopa Troop with unparalleled determination. He unleashed a barrage of star-infused energy, illuminating the battlefield and striking fear into the hearts of the enemy.
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)