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© h.hofstede (h.hofstede@hogeland.nl) |
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| 1. | a. | f(x) =
-4x2 + px + 2p f '(x) = -8x + p = 0 p = 8x f(x) = -4x2 + 8x · x + 16x f(x) = 4x2 + 16x |
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| b. | f(x) = x3
- 12px -
15 f '(x) = 3x2 - 12p = 0 12p = 3x2 p = 1/4x2 f(x) = x3 - 3x3 - 15 f(x) = -2x3 - 15 |
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| c. | f(x) = 2px2
+ x -
p f '(x) = 4px + 1 = 0 4px = -1 p = -1/(4x) f(x) = -2/(4x) · x2 + x + 1/(4x) f(x) = 1/2x + 1/(4x) |
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| 2. |
 |
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| Dat is nul als de teller
nul is: 2x(2 - px)
+ px2 = 0 4x - 2px2 + px2 = 0 4 - px2 = 0 px2 = 4 p = 4/x² |
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| 3. | y = ax2
+ bx + 8 y '= 2ax + b = 0 b = -2ax y = ax2 - 2ax2 + 8 y = -ax2 + 8 en die moet door (4, 20) gaan 20 = -a · 16 + 8 a = -3/4 b = -2ax = -6 |
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| 4. | a. |
8x2√x
-
px2 = 0 x2(8Öx - p) = 0 x = 0 ∨ 8Öx = p x = 0 ∨ Öx = 1/8 · p x = 0 ∨ x = 1/64p2 |
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| b. |
f(x)
= 8x2√x
-
px2 f '(x) = 20x√x - 2px = 0 2x(10√x - p) = 0 p = 10√x invullen in f: f(x) = 8x2√x - 10√x∙x2 = -2x2√x |
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| c. |
f(x)
= 8x2√x
-
10x2 f '(x) = 20x√x - 20x f '(4) = 160 - 80 = 80 y = 80x + b moet door (4, 96) gaan dus b = -224 |
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| 5. | a. |
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| f '(2) = 11/25
= 0,44 Raakpunt (2, -0.4) geeft b = 2,176 De raaklijn is y = 0,44x + 2,176 |
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| b. |
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| 3 -
px2 + 2px2 = 0 px2 = -3 p = -3/x² f(x) = x/ (3 + 3) Het is de lijn y = 1/6 ·x |
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| c. | Top bij y =
0,5 geeft 0,5 = 1/6
· x dus
x = 3 De grafiek gaat door (3, 1/2) 1/2 = 3/(3 - 9p ) 3 - 9p = 6 p = -1/3 |
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