Curve Sketching (1 Viewer)
- Thread starter Sunyata
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SeCKSiiMiNh
i'm a fireball in bed
1a) sketch 1/x^2 - 1 and x^3 then multiply the ordinates
nonsenseTM
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wtf, you're doing hsc in 2013
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Q2
a)
P(x) = x5 - 5cx + 1
P'(x) = 5x4 - 5c
Solving P'(x) = 0
=> x4 = c
If c < 0 then turning points do not exist, also note that if c < 0 then
- 5c > 0 and 5x4 ≥ 0
.: 5x4 - 5c > 0
This means P(x) is monotonically increasing for all real x
Since a polynomial of odd degree must always have at least one real root and here there are no turning points in the natural domain, there must be only one real root. (Draw a sketch to see this)
b)
Suppose turning points do exist and c > 0 then
x = - c1/4 or x = c1/4
If P(x) has three distinct zeros these turning points must have y-values opposite in sign (draw diagram to see this), hence
P(- c1/4).P(c1/4) < 0
[- c5/4 + 5c5/4 + 1][c5/4 - 5c5/4 + 1] < 0
(1 + 4c5/4)(1 - 4c5/4) < 0
1 - 16c5/2 < 0
c5/2 > 1/16
c > 1/162/5
c > (2-4)2/5
.: c > 2-8/5
Q3
a)
Let P(x) = x4 / 4 + x³ / 3 + x² / 2 + x + c
P'(x) = x³ + x² + x + 1
If P'(x) = 0
x³ + x² + x + 1 = 0
(x + 1)(x² + 1) = 0
=> x = - 1 for real x
P(-1) = 1/4 - 1/3 + 1/2 - 1 + c
= c - 7/12
For there to be no roots the polynomial should have all y-values > 0. This means that the sole turning point must have a y-value that is positive such that the polynomial is positive definite.
.: P(-1) > 0
=> c > 7/12
b) Question is carelessly worded. A polynomial of degree 4 will always 4 roots whether they be real or not. I think the question is asking for how many real roots.
If c ≤ 7/12 then P(-1) ≤ 0
This means the y-value of the turning point is non-positive. Since there is only one turning point (which can be shown to be the absolute minimum), then there must be two real roots when c < 7/12 and one repeated real root when c = 7/12
a)
P(x) = x5 - 5cx + 1
P'(x) = 5x4 - 5c
Solving P'(x) = 0
=> x4 = c
If c < 0 then turning points do not exist, also note that if c < 0 then
- 5c > 0 and 5x4 ≥ 0
.: 5x4 - 5c > 0
This means P(x) is monotonically increasing for all real x
Since a polynomial of odd degree must always have at least one real root and here there are no turning points in the natural domain, there must be only one real root. (Draw a sketch to see this)
b)
Suppose turning points do exist and c > 0 then
x = - c1/4 or x = c1/4
If P(x) has three distinct zeros these turning points must have y-values opposite in sign (draw diagram to see this), hence
P(- c1/4).P(c1/4) < 0
[- c5/4 + 5c5/4 + 1][c5/4 - 5c5/4 + 1] < 0
(1 + 4c5/4)(1 - 4c5/4) < 0
1 - 16c5/2 < 0
c5/2 > 1/16
c > 1/162/5
c > (2-4)2/5
.: c > 2-8/5
Q3
a)
Let P(x) = x4 / 4 + x³ / 3 + x² / 2 + x + c
P'(x) = x³ + x² + x + 1
If P'(x) = 0
x³ + x² + x + 1 = 0
(x + 1)(x² + 1) = 0
=> x = - 1 for real x
P(-1) = 1/4 - 1/3 + 1/2 - 1 + c
= c - 7/12
For there to be no roots the polynomial should have all y-values > 0. This means that the sole turning point must have a y-value that is positive such that the polynomial is positive definite.
.: P(-1) > 0
=> c > 7/12
b) Question is carelessly worded. A polynomial of degree 4 will always 4 roots whether they be real or not. I think the question is asking for how many real roots.
If c ≤ 7/12 then P(-1) ≤ 0
This means the y-value of the turning point is non-positive. Since there is only one turning point (which can be shown to be the absolute minimum), then there must be two real roots when c < 7/12 and one repeated real root when c = 7/12
X
xwrathbringerx
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Hmmm How exactly would you do question 1 (b)???
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Notice that if you let y = k in the equation in (a) and rearrange you end up with the cubic polynomial in (b). This graphically means we are looking for the intersection points of y = k and y = x3 / (x2 - 1). So draw a horizontal line through different parts of the graph and determine how many points of intersection there are for different values of k. The number of points of intersection should indicate the number of solutions to the cubic polynomial