June 12, 20XX
Today marked a significant turning point in the ongoing struggle against the formidable Mecha Mario on this same battlefield. The Mushroom Kingdom military forces have been locked in a relentless battle with this mechanized menace for what feels like an eternity. One hundred brave soldiers, equipped with unwavering determination, have faced countless hardships while trying to combat the overwhelming power of Mecha Mario. However, hope emerged from the depths of our despair as one of the Mushroom Kingdom's military weapon was finally unlocked: Robo Mario.
The battlefield was charged with anticipation as the soldiers prepared for the ultimate showdown. The sky above us was filled with smoke and the echoes of clashing metal, foretelling the intensification of the conflict. We were all aware that this battle could be the turning point in the war against Mecha Mario.
Robo Mario, a colossal creation of advanced technology, stood tall and resolute. Its gleaming armor shimmered in the sunlight, exuding an air of invincibility. The soldiers couldn't help but be in awe of the sheer magnificence and power Robo Mario possessed.
As the battle commenced, the clash between Mecha Mario and Robo Mario was nothing short of epic. The earth shook beneath our feet as they exchanged blows, their mechanical limbs striking each other with incredible force. Sparks flew and thunderous sounds reverberated across the battlefield, punctuated by the cries of both machines.
The intensity of the battle only grew with each passing moment. Mecha Mario's cunning strategies and relentless assaults tested the limits of Robo Mario's strength and agility. The soldiers on the ground fought valiantly, supporting Robo Mario in any way they could, firing salvos of projectiles and providing tactical advice. The determination was to protect the Mushroom Kingdom and its inhabitants fueled their every action.
With every passing minute, it became evident that this was not just a battle of machines; it was a battle of wills, a test of courage and resilience. Robo Mario, standing as a symbol of hope for the Mushroom Kingdom, refused to yield to the overwhelming power of Mecha Mario. It relentlessly fought back, using its advanced weaponry and strategic prowess to match its adversary blow for blow.
As the battle raged on, the soldiers became infused with a renewed sense of purpose. They fought with unparalleled bravery, pushing themselves to their limits, inspired by the unwavering determination of Robo Mario. The battlefield had transformed into a chaotic symphony of metal, fire, darkness, and determination.
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)