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Excalidraw Data
Text Elements
inc
dec
inc
f’(x) >0
f’(x) < 0
f’(x)>0
f’(x) = 0
0^2 = 0 1^2 = 1 (inc) -2^2 = -4 (dec) (-inf, 0) dec (0, inf) inc
horizontal tangent line
vertical tangent line
3x^2 - 3x 3x(x-1)
3x(x-1) 3(0)(0-1) 3(1)(1-1) both 0s, x = 0 and x = 1
3(-1)(-1-1) = 6 3(2)(2-1) = 6 3(1/2)(1/2-1) = -.75
domain (-inf, inf)
0
1
-1
6
2
6
1/2
-.75
inc from (-inf, 0)u(1,inf) dec from (0,1)
(x^2-4)^(2/3)
chain rule
f = x^(2/3) g = x
f’(g) * g’
2/3(x^2-4)^(-1/3) * 2x 4x/(3(x^2-4)^1/3 f’x= 0 when 4x/4 = 0, so zero at 0 ( horizontal tangent ) f’x= is undefined when 3(x^2-4)^1/3 = 0, so zero at 2 and -2 ^ sharp or vertical tangent
0
-2
2
-3
(4(-3))/(3(-3^2-4)^(1/3) -12/3(5)^1/3 - negative number 4(-1)/3((-1)^2-4)^(1/3) -4 / 3(-3)^(1/3) - positive number 4(1) / (3((1)^2-4)^(1/3) = 4/3(-3)^1/3 - negative number 4(3) / 3((3)^2-4)^(1/3) = 12/3(9-4)^1/3) - positive number
-1
3
1
f inc (-2,0)u(2,inf) f dec (-inf,-2)u(0,2)
f’ = 2x-2 2(1)-2 = 0 no undefined
2(0)-2 = -2 2(2)-2 = 2
1
0
2
dec (-inf, 1) inc (1,inf)
inc (-inf, -2)(0,1) dec (-2,0)u(1,inf)
2(0) = 0 2(-1) = -2 2(.5) = 1
1^2 = 1 -1+2 = 1
-2+3 = 1 -2^2 = 4 not continuous at -2
P = R-C P = 7.2x-.001x^2-2.4x+.0002x^2 P = 4.8x-.0008x^2 P’ = 4.8-.0016x 0 = -.0016x + 4.8 -4.8 = -.0016x 3000 = x
-.0016(2000) + 4.8 =1.6 -.0016(4000) + 4.8 = -1.6
inc (0, 3000) dec (3000, 6000)
so it is increasing from 0 to 3000
2x^2+12x+20
4x+12
4(-3)+12 = 0
4(-4) + 12 = -4
4(-2) + 12 = 4
dec: (-inf, -3) inc (-3, inf)
2x^2-16x+35 4x-16
4(4)-16 = 0
4x-16 inc (4, inf) dec (-inf,4)
-3x^2 + 18x -15
-3x^2+18x-15 = 0 x^2 - 6x + 5 = 0 ( divide both sides by -3 (x-1)(x-5) = 0
1, 5 are critical values
-3(0)^2 + 18(0) - 15 = -15 -3(2)^2 + 18(2) - 15 = -3(4) + 36 -15 = -12 +21 = 9 -3(6)^2 + 18(6) - 15 = -3(36) + 108 - 15 = -108 + 93 = -15
inc(1,5) dec(-inf,1)u(5,inf)
(x-1)/(x-5)
f x-1 g x-5
f’ 1 g’ 1
(f’g-fg’)/g^2 ((1)(x-5)-(x-1)(1))/(x-5)^2 (x-5-x+1)/(x-5)^2 (-4)/(x-5)
None, -4 is never zero. though there is a VA at 5.
(-4)/(x-5)
-4/(4-5)^2 -4/(-1)^2 -4/1
-4/(6-5)^2 -4/(1)^2 -4/1
always decreasing, va at 5 dec: -inf, 5 5,inf
-x^3 +6x^2 -12x + 5 -3x^2 + 12x -12
-3x^2 + 12x -12 = 0 x^2 - 4x + 4 = 0 (x-2)^2 = 0 (2-2)^2 = 0
-3(1)^2 + 12(1) -12 = -3 -3(3)^2 + 12(3) -12 = -3(9) + 36 -12 = -27 + 36 - 12 = -3
horizontal slope at 2
dec (-inf, 2) (2,inf)
chain rule f’(g(x)) * g’(x)
f = x^2 g = -5x+7
f’ = 2x g’ = -5
2(-5x+7) * -5 -10(-5x+7) 50x-70
50x-70 = 0 50x = 70 x = 70/50
(-5(1)+7)^2 (-5+7)^2 2^2 4
(-5(2)+7)^2 (-10+7)^2 (-3)^2 6
-x^3 + 3x^2 - 3x + 3
-3x^2 + 6x - 3
-3x^2 + 6x -3 = 0 -x^2 + 2x -1 = 0 x^2 -2x + 1 = 0 (x-1)^2 = 0 x = 1
-3x^2 + 6x - 3
-3(0)^2 + 6(0) -3 = -3 -3(2)^2 + 6(2) -3 -3(4) + 12 - 3 -12 + 12 - 3 -3
-3(4)^2 + 6(4) -3 -3(16) + 24 -3 -48 + 21 -27
decreasing on -inf, inf
216.756x - 0.009x
0⇐x⇐18200
216.756 - .018x
216.756 -.018x = 0 216.756 = .018x (216.756)/.018 = x
12042 = x
critical point 12042
216.756 - .018(12041) 216.756 - 216.738 = +
216.756 - .018(12043) 216.756 - 216.774 = -
increasing (-inf, 12042) decreasing (12042, inf)
BUT 0⇐x⇐18200
increasing (0, 12042) decreasing (12042, 18200)
4x^3 + 36x^2 + 96x
12x^2 + 72x + 96
12x^2 + 72x + 96 = 0
12(x^2+6x+8) = 0
x^2 + 6x + 8 = 0
(x+2)(x+4) = 0
-2 and -4
12x^2 + 72x + 96
12(-1)^2 + 72(-1) + 96 12 - 72 + 96 36
12(-3)^2 + 72(-3) + 96 12(9) -216 + 96 108 -216 + 96 -12
12(-5)^2 + 72(-5) + 96 12(25) - 360 + 96 300 - 360 + 96 36
inc: (-inf,-4), (-2, inf) dec (-4, -2)
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