Points of IntersectionDate: 08/31/97 at 17:01:35 From: Lisa Wang Subject: Points of Intersection Dear Dr. Math: I'm a high school student taking calculus. I encountered this problem, which really made me think. Please help me. The problem is: Estimate all points of intersection of the graphs f(x) = x + sin x and g(x) = x^3 by using a calculator or computer. How do you know you have found them all? I estimated the points of intersection by using a graphing calculator to be (-1.317, -2.285), (0,0), and (1.317,2.285). But I got stuck on the second question. I thought about it and came up with a partial explanation. But I somehow think it's not reasonable. f(x) = x + sin x 1. f'(x) = 1 + cos x: This means that at any point (x, y), the slope = 1 + cox x. We can think of the slope as the rate f(x) is changing. g(x) = x^3 1. g'(x) = 3x^2: This means that any point (x, y) has a slope of 3x^2. This also indicates that g(x) is changing at a rate of 3x^2. When you compare the derivatives (or slopes) of the two functions (curves), you see that g(x) is always changing at a greater rate than f(x). The slope of f(x) is always between 1 and 2 (since it's 1 + cos x); while the slope of g(x) increases to infinity. Does points of intersection have anything to do with the derivative (or the slope of th curves)? Or are they related to acceleration (the second derivative)? I'm confused and I don't know if I'm going in the right direction. Date: 08/31/97 at 19:51:44 From: Doctor Anthony Subject: Re: Points of Intersection The three solutions you give are correct, but your reasoning on the slopes of the two curves is not correct. f'(x) = 1 + cos(x) and since cos(x) varies between -1 and +1 the slope of f(x) varies between 0 and 2. g'(x) = 3x^2 This is less than the slope of f(x) when x is near 0 and accounts for the fact that other intersections occur in addition to the (0,0) point. When x > 1.31716 the gradient of f(x) has a maximum value of 2 while gradient of g(x) is greater than 5.2047. It follows that no further intersections are possible. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 08/31/97 at 21:14:50 From: Lisa Wang Subject: Re: Points of Intersection Dear Dr. Math: Now I don't understand why you are looking at maximum value of the slope of f(x) when x > 1.31716 (which is 2) and the slope of g(x) to determine any other points of intersection. Don't you have to plug x > 1.31716 into f(x) and g(x) rather than f'(x) and g'(x)? Also, I know that g'(x) is less than f'(x) when x is near 0. But I don't see how the different slopes determine points of intersection near 0. The values that I got for f'(x) at (0,0) and g'(x) at (0,0) are different. So I think I can draw the conclusion that the slopes of f(x) and g(x) at their intersection points are not equal. I don't quite understand how the slopes of functions affect their points of intersection. Date: 09/01/97 at 06:11:30 From: Doctor Anthony Subject: Re: Points of Intersection Look at the point of intersection where x = 1.31716. The y values of the two functions are equal at this point (since the curves cut here). We know that the slope of one of the curves (g(x)) is thereafter (as x increases) always greater than the slope of the other curve (f(x)), so there is no way that any more intersections can take place. We can conclude there will be no further intersections with x > 1.31716 At the point (0,0) the slope of g(x) is zero and the slope of f(x) is 2, so f(x) rises above g(x). However as x increases the slopes change rapidly, with g(x) increasing and f(x) decreasing, and this change leads to the curves crossing again at x = 1.31716 Thereafter the slopes determine that no further crossing points are possible, as described above. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 09/02/97 at 00:30:38 From: Lisa Wang Subject: Re: Points of Intersection Dear Dr. Math: I still don't quite understand the effect of slope on intersection points. My questions are: 1. At the points of intersection (0, 0), (1.31716, 2.28517), and (-1.31716, -2.285179), the slopes of f(x) and g(x) are not equal. How does slope determine points of intersection? In other words, when two functions intersect, what must their slopes be? (equal? greater? smaller?) 2. You said that "the slope of one of the curves (g(x)) is thereafter (as x increases) always greater than the slope of the other curve (f(x)), so there is no way that any more intersections can take place when x > 1.31716." I understand this. However, at the point of intersection (1.31716, 2.28517), the slope of g(x) is already greater than the slope of f(x). Why do you say that if the slope of g(x) is greater than the slope of f(x), then no intersection is possible? (since the two curves intersected when x = 1.31716 and at that point the slope of g(x) is greater than the slope of f(x)) Date: 09/02/97 at 05:37:11 From: Doctor Anthony Subject: Re: Points of Intersection At the point of intersection the slopes will in general be different, unless the curves are tangential to each other. It is what happens AFTER the point of intersection that interests us. If the curves diverge then NO FURTHER points of intersection can occur. >Why do you say that if the slope of g(x) is greater than the slope of >f(x), then no intersection is possible? (since the two curves >intersected when x = 1.31716 and at that point the slope of g(x) is >greater than the slope of f(x)) I think we have run into a language problem here. We had accepted that the curves intersected in three points (0,0) and (-1.31716, -2.28517), (1.31716, 2.28517). The question asked us to show that there were no FURTHER points of intersection. It is in pursuit of establishing this fact that I have been talking about the slopes of the two curves. Obviously at the three points of intersection already established we are not concerned about the slopes, except in so far as the slopes determine whether FURTHER points of intersection are possible. At the point (1.31716, 2.28517) the two curves intersect, but now as x increases one curve slopes up at a rate that is always greater than the other curve. Because of this the two curves can not cross each other AGAIN. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/