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x^2-x-4 2x-1 2(1/2)-1 = 0 critical point 1/2 2(0) - 1 = -1 2(1) - 1 = 1 dec(-inf, 1/2) inc(1/2,inf)
max and mins
relative to it’s surroundings, the largest y value
relative to it’s surroundings, smallest y value
x^4-x^3 4x^3-3x
4x^3 -3x^2 = 0 x^2(4x-3) = 0
x^2 = 0 0^2 = 0
4x-3 = 0 4x = 3 x = 3/4 critical points: 0, and 3/4
-1^2(4(-1)-3) -1^2(-4-3) 1(-4-3) negative
.5^2(4(.5)-3) 1/4(-1) negative
1^2(4(1)-3) 4-3 1 positive
dec: -inf, 3/4 inc: 3/4, inf
0 negative → negative nothing 3/4 negative → positive relative minimum
min(3/4, f(3/4)) (3/4)^4-(3/4)^3 -27/256 min(3/4,-27/256)
(x^4+1)/x^2 f = x^4 + 1 g = x
f’ = 4x^3 g’ = 2x
((4x^3)(x^2)-(x^4+1)(2x))/(x^2)^2 (4x^5-2x^5-2x)/x^4 2x(2x^4-x^4-1)/x^4) 2x(x^4-1)/x^4 2(x^4-1)/x^3
undefined at 0, because VA (Denominator = 0) undefined is when denominator = 0 horizontal tangent line is when f’(x) = 0 2(x^4-1) = 0 x^4-1 = 0 x^4 = 1 4rt(x) = 4rt(1) x = +/- 1
critical points -1, 1 va at 0
2((-16)^4-1)/(-2)^3) 2(15)/-8 30/-8 negative
2((-1/2)^4-1)/(-1/2^3) positive
2((1/2)^4 - 1)/(1/2^3) negative
-1 and 1 are minimums min -1,2 1,2 f(1) = 2 f(-1) = 2
f x g x+1
f’ 1 g’ 1
1(x+1) - x(1) / (x+1)^2 x + 1 - x / (x+1)^2 … “let’s move on to 3.6”
bro we never did an example ourselves what
3.6
find critical numbers plug critical numbers and endpoints into f
x^2-6x+2 2x-6
2x-6 = 0 2x = 6 x = 3
3^2-6(3)+2 = 9 - 18 + 2 = -9 + 2 = -7 0^2-6(0)+2 = 2 5^2-6(5)+2 = 25 - 30 + 2 = -5 + 2 = -3
absolute max: 0,2 absolute min: 3,-7
3x^4 -4x^3 12x^3 -12x^2 12x^2(x-1)
(1-1)=0 12(0)^2 = 0
critical value 0, 1 endpoints -1 and 2
(3)-1^4 -4(-1)^3 3 -4(-1) 3 + 4 7
(3)0^4 -4(0)^3 0
(3)1^4 -4(1)^3 3 -4 -1
(3)2^4 -4(2)^3 48 -4(8) 48 -32 16
absolute max 2,16 absolute min 1,-1
2x-3x^(2/3) 2-2x^(-1/3) 2(1-1/x^3
7x^2-14x+6
14x-14
14x-14 = 0 14x = 14 x = 1
14x-14
14(0)-14 -14 ( dec )
14(2) - 14 (inc)
dec → inc local min
7x^2-14x+6 7(1)^2-14(1)+6 7-14+6 -7+6 -1
f/g (64x^2 + 1) / x
f = 64x^2 + 1 g = x
f’ = 128x g’ = 1
128x^2 - (64x^2 +1) / x^2 64x^2 - 1 / x
64x^2-1/x^2 = 0 64x^2 - 1 = 0 64x^2 = 1 x^2 = 1/64 x = sqrt(1/64) x = +- 1/8
0 is VA
1/8, -1/8, are critical points
(64x^2-1)/x
-1 64(-1)^2-1/-1^2 64-1 63
-1/8
.1 64(.01)-1 / .01 .64-1 /.01 -.36/.01 -36
1/8
1 64(1)-1 / 1 64-1 63
64(1/8)^2 + 1 )/(1/8) 64(1/64)+1 / 1/8 1+1 / (1/8) 16
64(-1/8)^2 + 1) /(-1/8) 64(1/64) + 1)/(-1/8) 1 + 1 / (-1/8) 2/(-1/8) -16
x + 81/x x + 81x
1 - 81x
1 - 81x^-2 = 0 1 - 81/(x^2) = 0 -81/(x^2) = -1 -81 = -x^2 81 = x^2 +-9 = x
9, -9
1 - 81/(x^2)
-10 1 - 81/(-10^2) 1 - 81/100 = 19/100
-9 ( local maxima at -9 )
1 1 - 81/(1^2) 1 - 81 -82
9 ( local minima at 9 )
10 1 - 81/(10^2) 1 - 81/100 = 19/100
-9 + 81/-9 -9 - 81/9 -9 - 9 -18
9 + 81/9 9 + 9 18
2x^(4/3) - 8x + 10
(8/3)x^(1/3) - 8
(8/3)x^(1/3) - 8 = 0 (8/3)x^(1/3) = 8 x^(1/3) = 8/(8/3) cbrt(x) = 8/(8/3) x = (8/(8/3))^(3) x = 27
(8/3)x^(1/3) - 8
26 (8/3)26^(1/3) - 8 -.1
27
28 (8/3)28^(1/3) - 8 .09
neg pos local min at 27
2(27)^(4/3) - 8(27) + 10 2(81) - 8(27) + 10 162 - 8(27) + 10 162 - 216 + 10 162 - 206 -44
(27,-44)
16x^2 - 1 ) / x
f = 16x^2-1 g = x
f’ = 32x g’ = 1
f’g-fg’ / g
32x^2-(16x^2-1) 32x^2-16x^2+1 16x^2+1 / x
16x^2 + 1 / x^2 = 0 16x^2 + 1 = 0 16x^2 = -1 x^2 = -1/16
None
23 + (343t^2 )/ (t^2+93)
f= 343t^2 g= t^2+93
f’ = 686t g’= 2t
f’g - fg’ / g
686t(t^2+93) - (343t^2)(2t) / (t^2+93)^2 686t^3 + 63798t - 686t^3 / (t^2+93)^2 63798t / (t^2+93)
va at 0
numerator and denominator will always be positive >0, which problem implies, so can never divide to get a negative number, so always positive from 0,inf
343t^2
t^2 + 93
343
1 + 93/t
as we approach infinite t value 343
1 + 0
343/ 1 = 343
343 + 23 = 366
4+(7t^2)/(t^2+210)
7t^2 / (t^2+210)
f = 7t^2 g = t^2 + 210
f’ = 14t g’ = 2t
f’g - fg’ / g
14t(t^2+210) - 7t^2(2t) / (t^2+210)^2 14t^3 + 2940t - 14t^3 / (t^2+210)^2 2940t / (t^2+210)
0,inf increasing?
^ yes. i was right, time cant be negative, and both denominator and numerator are always gonna be positive if input t is positive
-8x^2 + 16x + 3
-16x + 16
-16x + 16 = 0 -16x = -16 16x = 16 x = 1
-6
0 -16(0) + 16 16
1
2 -16(2) + 16 -32 + 16 -16
4
-8(1)^2 + 16(1) + 3 -8 + 16 + 3 -8 + 19 11
-8(-6)^2 + 16(-6) + 3 -8(36) -96 +3 -288 -96 + 3 -381
-8(4)^2 + 16(4) + 3 -8(16) + 16(4) + 3 -128 + 64 + 3 -61
3x^2 -24x
6x-24
6x - 24 = 0 6x = 24 x = 4
3 6(3)-24 = -6
4
5 6(5)-24 = 6
3(0)^2-24(0) 0
3(4)^2-24(4) 3(16) - 96 48-96 -48
3(7)^2-24(7) 3(49) -168 147-168 = -21
-x^3 - 6
-3x
-3x^2 = 0 x^2 = 0 0
-1 -3(-1)^2 -3(1) -3
0
1 -3(1)^2 -3
-4 -(-4)^3-6 64 -6 58
0 -(0)^3-6 -6
5 -(5)^3-6 -125-6 -131
abs max, -4, 58 abs min 5, -131
3x^3 -36x 9x^2 - 36
9x^2-36 = 0 9x^2 = 36 x^2 = 4 x = 2
-3 3(-3)^3 - 36(-3) 3(-27) + 108 -81 + 108 27
2 3(2)^3 -36(2) 3(8) -72 24-72 -48
7 3(7)^3 -36(7) 3(343) - 252 1029 - 252 777
min 2,-48 max 7,777
2x^3 -15x^2 -36x
6x^2 -30x -36
6x^2 - 30x - 36 = 0 x^2 - 5x - 6 = 0 (x+1)(x-6) = 0 -1 and 6 are critical points
-7 2(-7)^3-15(-7)^2-36(-7) 2(-343) -15(49) +252 -686 -735 + 252 -1169
-1 2(-1)^3-15(-1)^2-36(-1) 2(-1)-15+36 -2-15+36 19
6 2(6)^3-15(6)^2-36(6) 2(216) -540 -216 432-540-216 -324
9 2(9)^3 -15(9)^2-36(9) 2(729) -15(81) -324 1458 - 1215 -324 -81
min -7,-1169 max -1, 19
(5x+1)(2x-3)
f 5x+1 g 2x-3
f’ 5 g’ 2
5(2x-3) * 2(5x+1) 10x-15 + 10x + 2 20x - 13
20(1) - 13 20 - 13 7
4x^(1/4) - (qdrt(12)/12)x x^(-3/4) - qdrt(12)/12
x^(-3/4) - qdrt(12)/12 = 0 x^(-3/4) = qdrt(12)/12 12x^(-3/4) = qdrt(12) 12x^(-3/4) = qdrt(12) 12/x^(3/4) = qdrt(12) 12^4 / x^(3/4)^4 = qdrt(12)^4 20736 / x^3 = 12 20736 = 12x^3 1728 = x^3 cbrt(1728) = x 12 = x
critical point at 12
8 4(8)^(1/4) - (qdrt(12)/12)(8) 5.48636
12 4(12)^(1/4) - (qdrt(12)/12)(12) 5.58362
26 4.9997
30x-3x
30-6x
30 = 6x 5 = x
critical point at 5
0 30(0)-3(0)^2 0
5 30(5)-3(5)^2 150-3(25) 150-75 75
21 30(21)-3(21)^2 630-3(441) 630-1323 -693 maximum 5, they should sell 5.
-8x+7 -8
-8 -8(-8) + 7 64 + 7 71
6 -8(6) + 7 -48 + 7 -41
94x-.05x^2 - .04x^2 - 22x - 800 72x -.09x^2 - 800
72 -.18x 72 - .18x = 0 72 = .18x 400=x
400
there is some other stuff i could do here to check break even points but profit is positive at 400 so who gaf
(x^2-1)^(1/5)
f = x^(1/5) g = x^2 - 1
f’ = 1/5x^(-4/5) g’ = 2x
1/5(x^2-1)^(-4/5) * 2x
1/5(x^2-1)^(-4/5) * 2x = 0
2x/(5(x^2-1)^(4/5)) = 0
critical point at 0
va at 1, -1
-2 ((-2)^2-1)^(1/5) 1.2457
-1 ((-1)^2-1)^(1/5) 0
0 ((0)^2-1)^(1/5) -1
1 ((1)^2-1)^(1/5) 0
4 ((4)^2-1)^(1/5) 1.7187
x^2/120 - 7x/15 + 4427/120
x^2/120 - 56x/120 + 4427/120
(x^2-56x+4427)/120
f = x^2-56x+4427 g = 120
f’ = 2x-56 g’ = 0
(2x-56)(120)/120
(2x-56)/120
2x-56/120 = 0 2x-56 = 0 2x = 56 x = 28
0 0/120 - 7(0)/15 + 4427/120 36.8916
28 30.3583
80 52.8916
28, 30.358
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