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g(x) = 1/7x^7 - x
find intervals of increasing and decreasing
find relative extrema
1/7x^7 - x
x^6 - 1
x^6 - 1 = 0 1^6 - 1 = 0 -1^6 - 1 = 0
critical points 1 and -1
-2^6 - 1 = 63
0^6 - 1 = -1
2^6 - 1 = 63
inc dec inc
local max at -1 local min at 1
1/7(-1) + 1
1/7(1) - 1
-1/7 + 1 = 6/7
1/7 - 1 -6/7
max at -1,6/7 min at 1, -6/7
wanna know when the slope of the tangent line is increasing when is tangent line getting steeper / less steep.
were gonna take derivative again gonna call it concava when it opens up
concave up
concave down
second derivative
x^3 +5x^2 3x^2 + 10x 6x + 10
concave up when f’ is increasing which occurs when f” is positive.
concvae down when f’ is decreasing which occurs when f” is negative
concave down from -inf, 0
concave up from 0, inf
-2x^3 - 8x^2 + 4
-6x^2 - 16x
-12x - 16
-12x - 16 = 0
12x = 16 x = 16/12 x = -4/3
-12(-1) - 16 = -4
-12(-2) - 16 = 8
concave up from -inf, -4/3
concave down from -4/3, inf
-4x^3 - 6x^2 + 1 -12x^2 - 12x -24x -12 -24x -12 = 0 -24x = -12 x = -12/24 x = -1/2
-24(-1) - 12 = 12
-24(0) - 12 = -12
concave up from inf, -1/2 concave down from -1/2, inf
x^4 + x^ 3 - 3x^2 + 1 4x^3 + 3x^2 -6x 12x^2 + 6x - 6
12x^2 + 6x - 6 = 0
6(2x^2 + x - 1) = 0
2x^2 + x - 1 = 0
factors that multiple to -2 and add to 1 are 2 and -1
2x^2 + 2x - x - 1 ( factored )
(2x-1)(x+1)
(2x-1)(x+1) = 0
2x-1 = 0
2x = 1
x = 1/2
x+1 = 0 x = -1
-2
-1
0
1/2
1
6(2(-2)^2 -2 -1) 6(2(4) -2 - 1) 6(8 - 3) 6(5) 30
6(2(0)^2 + 0 -1) 6(-1) -6
6(2(1)^2 + 1 - 1) 6(2) 12
concave up from -inf, -1 and 1/2, inf
concave down from -1, 1/2
points of inflection -1, -2 1/2, 7/16
-1/10x^3 + 6x^2 + 400
-3/10x^2 + 12x -6/10x + 12
-6/10x + 12 = 0 -6/10x = -12 x = 20
1 -3/5(10) + 12 = pos num 30 -3/5(30) + 12 = neg num
after 20 it becomes concave down.
20
first derivative test is number line thing
we are now testing for concave up or concave down at inflection points.
x
3x^2 - 3
3x^2 - 3 = 0 3x^2 = 3 x^2 = 1 x = 1 or -1
6(1) = 6 6(-1) = -6
minimum at 1 maximum at -1
-3x^5 + 5x^3 -15x^4 + 15x^2 -60x^3 + 15x
-15x^4 + 15x^2 -15x^2(x^2-1)
-15x^2 = 0 x^2 = 0 sqrt(x^2) = 0 x = 0
x^2-1 = 0 x^2 = 1 x = +-1
-60(-1)^3 + 30(-1) = 30 > 0 concave up min at -1, f(-1) -60(0)^3 + 30(0) = 0 == 0 linear -60(1)^3 + 30(1) = -30 < 0 concave down max at 1, f(1)
5x^3 - 120x^2 - 10x - 9 15x^2 - 240x - 10 30x - 240
30x-240 = 0 30x = 240 x = 8
30(7) - 240 = -
30(9) - 240 = +
concave up from 8,inf concave down -inf, 8
5(8)^3 - 120(8)^2 - 10(8) - 9 -5209
4x^3-4x^2-6x 12x^2 -8x - 6 24x - 8
chain rule f = x^(1/2) g = 9x+7
f’ = 1/2x^(-1/2) g’ = 9 f’(g(x)) * g’(x)
(1/2)(9x+7)^(-1/2) * 9 (9/2)(9x+7)^(-1/2)
f = (9/2)x^(-1/2) g = 9x+7
f’ = -9/4x^(-3/2) g’ = 9
(-9/4)(9)(9x+7)^(-3/2) (-81/4)(9x+7)^(-3/2) -81/(4(9x+7)^(3/2)
f(6) = -24 f’(x) != 0 f”(x) = 0
f(x) = x - 30
-2x^3+15x^2-24x+24 -6x^2+30x-24 -12x+30
-2x^3+15x^2-24x+24 -6x^2+30x-24 -12x+30
-6x^2+30x-24 -6x^2+30x-24 = 0 x^2 -5x + 4 = 0 (x-1)(x-4) critical points at x=1 and x=4
-12(1)+30 -12+30 18
concave up. minimum at 1
-12(4)+30 -48+30 -18
concave down. maximum at 4.
-2(1)^3+15(1)^2-24(1)+24 13
-2*(4)^3+15*(4)^2-24*(4)+24 40
4x^2+16x-48 8x+16 8
test doesnt apply? i think? or whole thing is concave up.
8x+16=0 8x=-16 x=-16/8 x=-2
4(-2)^2+16(-2)-48 -64
minimum at -2, -64, i think. no local max. ^ yes this was right.
-x^2+2x+24 -2x+2 -2
-x^2+2x+24 -2x+2 -2
concave down whole time
-2x+2=0 -2x=-2 x=1
maximum at 1,f(1) -(1)^2+2*(1)+24 25
x^4-12x^3-5 4x^3-36x^2 12x
4x^3-36x^2=0 4(x^3-9x^2)=0 x^3-9x^2 = 0 (x^2)(x-9)=0 x = 0 and x = 9
12x^2-72x 12(0)^2-72(0) = 0 test doesnt apply
0^4-12(0)^3-5 -5
9^4-12(9)^3-5 -2192
4(-1)^3-36(-1)^2 -40
0
4(5)^3-36(5)^2 -400
9
4(10)^3-36(10)^2 400
0 is neg neg, not a max or min
9 is a neg pos, min at 9.
6x^3-54x^2+90x 18x^2-108x+90 36x-108
18x^2-108x+90=0 18(x^2-6x+5)=0 x^2-6x+5 = 0 (x-1)(x-5) = 0 x=1 and x = 5
36(1)-108 =-72 36(5)-108=72
concave down at 1 concave up at 5
max at 1,f(1) min at 5,f(5)
6(1)^3-54(1)^2+90(1) 42
6(5)^3-54(5)^2+90(5) -150
x^2-4x^(1/2)-2 2x-2x^(-1/2) 2+x^(-3/2)
x^2-4x^(1/2)-2 2x-2x^(-1/2) 2+x^(-3/2)
2x-2x^(-1/2) = 0 2x = 2x^(-1/2) x = x^(-1/2) x = 1
2+1^(-3/2) 3
concave up at 1 min at 1
no max
1
1-4(1)-2 -5
x^4-32x^2-10 4x^3-64x 12x
4x^3-64x=0 x^3-16x=0 x^3=16x x^2=16 x=+-4 OR 0
12(-4)^2-64 12(16)-64 128
concave up minimum at -4
12(4)^2-64 12(16)-64 128
concave up minimum at 4
12(0)^2 - 64 -64 concave down maximum at 0
4^4-32(4)^2-10 -266
-4^4-32(-4)^2-10 -266
0^4-32(0)^2-10 -10
chain rule f x^2 g 3x^2+25
f’ 2x g’ 6x
2(3x^2+25) * 6x 12x(3x^2+25)
then product rule f’g + fg’
f 12x g 3x^2+25 f’ 12 g’ 6x
12(3x^2+25) + 12x(6x) 36x^2+300 + 72x^2 108x^2+300
chain rule f x^2 g 3x^2+25
f’ 2x g’ 6x
2(3x^2+25) * 6x 12x(3x^2+25)
then product rule f’g + fg’
f 12x g 3x^2+25 f’ 12 g’ 6x
12(3x^2+25) + 12x(6x) 36x^2+300 + 72x^2 108x^2+300
12x(3x^2+25) = 0 36x^3 + 300x = 0 6x^3 + 50x = 0 0 is a critical value 6x^2+50= 0 6x^2 = -50 x^2 = -50/6 x = sqrt(-50/6) not gonna come near 0
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