We started off today’s class by going over Friday’s Quiz. I think the first question was quite straightforward so I won’t get into that.
2) Use linear approximation to find the value of (1.001)^100. I had trouble understanding this question, but after Mr.K explained it, it made much more sense. The important thing to recognixe is that you have a value very close to one that would be easy to solve for.
1) Find a function to represent the question. We also find the derivative of the function, so that we can find the equation of the line later.
2) Choose a simple number close to the value we’re approximating. In this case, 1 is the easiest value. By finding the f(1) and f ‘(1), we can use the point-slope formula to find the tangent line.
3)Now we can plug in the value we’re approximating, 1.001, and we get an answer of 1.1, which is a good approximation of the actual value, 1.1051.
3) Using 3 iterations of Newton’s Method, by hand, find the root of f(x)= x^5-x+1.
Start with x= -1.
Newton’s method uses progressively more accurate tangent lines to approximate the root of a function. First of all we need to have f(x) and f ‘(x) so that we can find the tangent line.
The second step shows the familiar point-slope formula. Because the root is where y = 0, we can get rid of y, and then rearrange the equation to solve for the next x-value.
The first iteration is done by finding f(-1) and f ‘(-1). Then plugging these into the rearranged formula, we have the next x-value. We repeat the process using this next value instead of x=-1, and so on until reaching however accurate a solution we want.
From here we moved on to the lesson itself, all about finding horizontal and vertical asymptotes and using derivatives to find out what's happening around the asymptotes.
1) Draw a graph of f(x)= 1/(x-3)
Because the limit of the function as x approaches 3 is one divided by zero (undefined), we know that there is a vertical asymptote at x=3. If the limit were zero divided by zero, it would be indeterminate (impossible to know what is happening).
By finding where the function is increasing/decreasing and concave up/down, we can determine what the graph is doing on either side of the asymptote. We know that there are no roots because the numerator always exists. So we get a graph like this:
2) Draw a graph of g(x)= 1/(x^2+1)
Once again, the numerator always exists so there are no roots. And there are no vertical asymptotes because the denominator will always exist. By finding the limit as x approaches infinity, we can determine that there is a horizontal asymptote at y=0.
Technically, we should also find the limit as x approaches negative infinity, in case the graph is doing something else on the other side of the x-axis, but this is a nice happy function so we don’t bother.
From here it’s more or less the same as the previous question. Finding g’ and g”, we learn key information about the graph that allows us to draw it. For example, we know there’s a max at (0,1) because g’ changes from positive to negative at x+0.
This graph is a very well-known one in mathmatics, called the 'Witch of Agnesi". To find out why, look here: http://mathworld.wolfram.com/WitchofAgnesi.html
3) Next we looked at (2x^2)/(x^2-5x+6).
If we try to find the limit as x approaches infinity, we will get ∞/∞, which is indeterminate. In this case, to find the horizontal asymptote we divide each term by the highest power of x in the function. There are three possible outcomes when we do this:
1) If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at y=0.
2) If the degree of the numerator is greater than the degree of the denominator, the horizontal asymptote does not exist.
3) If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the leading coefficient of the numerator divided by the leading coefficient of the denominator.
It is probably easier to memorize these three rules, but if you think you will forget them over time, it is probably better to remember that dividing by the highest power of x will allow you to find the horizontal asymptote. That's what Sherlock Holmes would do, anyway. :)
In this problem, the third case applies. So we get an asymptote at 2/1.
If the numerator had been 2x^3, the second case would apply and the horizontal asymptote wouldn’t exist.
And if the x^2 in the denominator had been x^3, then the horizontal asymptote would have been at y=0.
So that’s about it. Just remember...
By finding the limit as x approaches infinity you can find any horizontal asymptotes.
If the limit as x approaches some value a approaches infinity, there's a vertical asymptote.
And Jupiter has 63 known moons.
Next scribe shall be… Anh.
3 comments:
I say hall of famer!!
...and I'll tell you why...
First, because these lessons are fairly new. It's hard to do a scribe post on something you just learned about, and suzanne did an excellent job at it.
Second, because of there are any errors on any of Suzanne's posts, I haven't found them... and that tells me how much effort and attention to detail she puts in!
So, I think this should be inducted into the hall of fame.
I made an error above.. I meant to say "Second, because IF there are any...."
I would also like to see Suzanne's post in the hall of fame, because of the great amount of work she put into it. She had a lot of things to cover, and she didn't take the easy way out of things.... she deserves the recognition for all the effort and hard work. Good job Suzanne =)
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