June 16, 20XX
Today has been an unimaginable day filled with chaos and destruction as the Battle of Mushroom Raceways reached new levels of intensity. The clash between good and evil has escalated to a point where even the bravest of hearts tremble in fear. Mecha Sonic, a formidable foe, continues to engage in a relentless battle against our heroes: Mario, Sonic, Luigi, Shadow, and Yoshi. The clash of titans is a sight to behold, but it is also a reminder of the immense danger that engulfs us.
As the battle rages on, Mecha Mario and Robo Mario engage in a fierce duel of their own. The air is filled with the clashing of metal and the crackling of energy beams as these mechanical counterparts fight for dominance. It is a surreal spectacle to witness, reminding us of the sacrifices made in the pursuit of justice.
In the midst of this chaotic scene, a glimmer of hope emerges in the form of Prince Mallow of Nimbus. He embodies the noblest of virtues, providing much-needed medical supplies and assistance to the wounded soldiers and civilians caught in the crossfire. The Prince's selfless acts of kindness offer solace to those suffering amidst the destruction, instilling a sense of unity and determination in our ranks.
The battlefield has transformed into a perilous landscape, littered with debris and scorched earth. The once vibrant Mushroom Raceways are now a battleground, bearing witness to the clash between good and evil. The air is heavy with tension, and the echoes of battle cries mingle with the cries of pain from the wounded.
It is in moments like these that we are reminded of the fragility of life and the true cost of war. Lives hang in the balance, and the weight of our choices becomes ever more apparent. The bravery and unwavering resolve of our heroes inspire us to fight on, to protect the innocent and restore peace to Mushroom Kingdom and Mobius.
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)