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Don't have my book or audio? You can download a free sample of my book and audio lectures containing Lessons 1 and 2:
Did you read my tips on how to study and learn Math 1500? If not, here is a link to those important suggestions:
These are tips for the second assignment in the Distance/Online Math 1500 course, but I strongly recommend that you do this assignment as homework even if you are taking the classroom lecture section of the course. These assignments are very good (and challenging)
practice. The first assignment is a great way to build and review key skills that will be helpful for this course. Here is a link to the actual assignment, in case you don't have it: You should thoroughly study Lesson 9: Curve-Sketching before attempting this lesson.
Be sure to compute and simplify the necessary derivatives and double-check you are right before you proceed to answer the
questions.
Make sure you read my tips on how to compute and simplify derivatives after question 4 in Lesson 9 (starts on page 276).
This question is asking for the critical
numbers. That means they want the critical points and singular points. The top and bottom zeros of the first derivative are the critical numbers. Make sure you give both the x and y coordinates of your critical numbers, even though it would be fine to just give the x values in this question (i.e. make a table of values for each). Recall, as I say in Lesson 9, e^u has no
zeros.
In part (c), you will get a negative exponent when you do the derivative. Pull that down to the denominator and then get a common denominator to properly identify the top and bottom zeros. You may find it easier to just guess at what the top and bottom zeros are. Be organized. Sub in x=0, 1, -1, 2, -2, ... There is no shame in just using trial and error to find zeros when things are difficult to
factor. Facts are facts. If you find an x-value that causes either a top or bottom zero, then there is no doubt. Hint: There are three critical numbers in part (c).
Similar to my Lesson 9, question 5. Make sure you include the sentences I box in in your answer as that is necessary to justify your conclusions.
This is a Mean Value Theorem question. Click the link below for the procedure to follow to "verify" the Mean Value Theorem: This question is pretty challenging.
Some hints: Pull all the terms over to the left side of the inequality so that you instead are trying to prove that x + cosx
- 1 is greater than or equal to 0. Let f(x) = x + cosx - 1.
Apply the Mean Value Theorem to that function like you did in question 3 above. Use [0, x] for your interval [a, b].
Note that sinx is never lower than -1 and never higher than 1. That will come in handy.
What this question is saying is that, for the given function, apparently the Mean Value Theorem does not work, because there is no c such that a < c < b (-1 < c < 1, in this case) where f'(c) = 0 as the Mean Value Theorem would
predict.
Consider the conditions f(x) must meet in order for Mean Value Theorem to apply. Is the function continuous on [a, b]? Is it differentiable on (a, b)? Which is to say, is the derivative defined for all values between a and b? Look for bottom zeros. A function or derivative is undefined if the bottom is zero.
If the function is not continuous, or not differentiable, then the Mean Value Theorem does not apply, and so it is no surprise if the Mean Value Theorem does not work.
You are analyzing f ' (x). The first derivative tells you where a function is increasing or decreasing and if the critical points are local maximums, local minimums, or neither. Be sure to give the (x, y) coordinates of
all the local extremes you identify.
Make sure you look at the tips on page 276 for assistance in simplifying your derivatives.
Remember my hint for 1(c) above when you analyze part (b).
You are analyzing f '' (x). The second derivative tells you where a function is concave up or concave down and if you have inflection points. Be sure to give the (x, y) coordinates of all the inflection points you
identify.
A classic curve sketch problem. Expect something similar on your final exam. Note that, when you have a vertical asymptote, there must be a curve drawn on both the left and right side of the vertical asymptote. If one side lacks any points of
interest, you can plot a "random" point to help orient yourself (just add a relevant x value to your table of values). Alternatively, you can also visualize what the curve in that section must look like by using your first and second derivative sign diagrams to guide you.
Take a look at my Practise Problem 1 for a similar example. Note that I choose a random point where x = -2 to put on my table of values just so that I can get a
feel for where the curve might be on the left side of the vertical asymptote at x = -1 because no points of interest occurred on that side of the asymptote.
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