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Hard Question (1 Viewer)
- Thread starter jjj
- Start date
Are you sure that you've been asked to find the solutions, and not simply to determine how many there are?
I can see that you could approximate the solutions (by Newton's method), but I don't see any easy way to actually solve that equation.
I can see that you could approximate the solutions (by Newton's method), but I don't see any easy way to actually solve that equation.
First off, you've got to have your graphs roughly to scale.
You need to be able to draw it pretty accurately.
That looks like one of the questions from the 2u Fitzpatrick book- exercise 17e, Q1-
If you take a look at the usual examples on page 393, you'll see that when they graphed it, by observation, they obtained a very rough estimate.
(Drop a perpendicular down from the point of intersection and by observation, guess where it lands, pretty much!)
From that, you just need to use your calculator to come up with a more precise answer.
Pretty dodgy way to do it, actually, but that's the way my school was taught to do it, but it works pretty well when youve got an accurate graph.
I don't think you'll be asked to do anything like that in an exam, so don't worry about it too much.
You need to be able to draw it pretty accurately.
That looks like one of the questions from the 2u Fitzpatrick book- exercise 17e, Q1-
If you take a look at the usual examples on page 393, you'll see that when they graphed it, by observation, they obtained a very rough estimate.
(Drop a perpendicular down from the point of intersection and by observation, guess where it lands, pretty much!)
From that, you just need to use your calculator to come up with a more precise answer.
Pretty dodgy way to do it, actually, but that's the way my school was taught to do it, but it works pretty well when youve got an accurate graph.
I don't think you'll be asked to do anything like that in an exam, so don't worry about it too much.
1. draw y = sin2x
2. draw y = x/2 on the same set of axes- (Important- it MUST be to scale).
3. You will observe 2 points of intersection if you drew it correctly- one at (0,0), and one between x = 0 and x = pi/2.
4. At the 2nd point of intersection, draw a line down from the point to the x axis such that the line is perpendicular to the x axis.
5. Estimate where that line cuts the x axis.
6. Use your calculator to check that point for its two values; that is:
y1 = (X/2)
y2 = sin2X
You're trying to find a point where y1 and y2 are the same.
Keep on subbing in values, from around the area that you think the approximate solution lies in.
Try to get as close as possible.
2. draw y = x/2 on the same set of axes- (Important- it MUST be to scale).
3. You will observe 2 points of intersection if you drew it correctly- one at (0,0), and one between x = 0 and x = pi/2.
4. At the 2nd point of intersection, draw a line down from the point to the x axis such that the line is perpendicular to the x axis.
5. Estimate where that line cuts the x axis.
6. Use your calculator to check that point for its two values; that is:
y1 = (X/2)
y2 = sin2X
You're trying to find a point where y1 and y2 are the same.
Keep on subbing in values, from around the area that you think the approximate solution lies in.
Try to get as close as possible.
Just to clarify, by 1/2x, do you mean (1 / 2) * x, ie x / 2, as Faera has used, or do you mean 1 / (2 * x),
ie (2x)^-1? If it's the first one, Faera's method will work, if it's the second, then the second graph needs to be the hyperbola y = 1 / (2x), not the line y = x / 2.
That was the point of my first post - to find the solutions you need to solve simultaneously - and I don't see a way to do that in this case. To determine how many solutions there are, or to approximate the solutions, you can use the graph, or a number of other approximation methods, like Newton's method.
