Advice on how to do well in class

This post will be added on to by all authors who are willing, and if you aren't an author and have a tip, feel free to make a comment.

1) Read every section the day before the class. This will help you be more on top of the lesson in class. It doesn't matter whether you're not doing well or if you're a prodigy, this will help you.

The Substitution Rule

The Substitution Rule is basically the reverse of the Chain Rule.

You find a sub-function within the function you are integrating, and represent it with a new variable. (The book, and Prof. Newberry too, uses u to represent the function).

Before you rewrite $g(f(x))dx  as  $u du , * make sure you go through and evaluate du:

du/dx = _    =>    du = _dx

If you're lucky, the value multiplied by dx when evaluating du will account for some of the things the function represented by u was multiplied by. Whether it does or doesn't, you will need to divide the remaining parts of the original function by the right side of the equation evaluating du to cancel out the multiplication by du.

To make this a bit clearer, I'll do Section 5.5 Example 2:

Find   Int = $ sqrt(2x+1) dx 

For u = 2x +1 ,

du/dx = 2   =>   du = 2 dx  => dx = du/2

Int = $ sqrt(u) du/2  =  1/2 $ u^(1/2) du

= (1/2) * u^(3/2) / (3/2)  + C

= (1/3) * u^(3/2)  + C

= (1/3) * (2x+1)^(3/2)  + C

If you took the trouble to read the above, you will have seen that in the first step of solving the integral using u, du was divided by 2 to correct for the fact that du/dx = 2 . 

I hope this helps.

*I don't have an integral symbol on my keyboard (who does?), so I'm using $ instead of the vertically stretched S.

Helpful Notes on Anti-Deriving

When taking a derivative of an ax^n equation, you first multiply the coefficient by the power, and then you subtract 1 from the exponent.

With integrals, it's the reverse. First you add 1 to the exponent, and then you divide the coefficient by the new exponent.

I have found this a little hard to remember in my math autopilot, but if this sequence can be memorized, it should be easier.

Keep in mind: write the x and its new exponent first, then figure out the coefficient.

God bless!

Slant Asymptotes

Ok, I have a question.

How do you figure out slant asymptotes?

I just was working on S 4.5 #21, use the guidelines to sketch y=(x^2+x-2)^(1/2)

It looked like there might be, there oughta be, some slant asymptotes, but the guidelines given in the book only work for rational functions, not radical functions.

Either I'm wrong that this has slant asymptotes, or there is some other way besides long division to figure out the equation of a slant asymptote. Or both.

Curve Sketching

This might be the hardest thing we've done yet. (Or maybe I'm just getting lazy all of a sudden. [lol])

So, the book says:

1st figure out the domain, the values of x for which the equation can be used.

2nd find the x- and y-intercepts

3rd figure out whether the function is even (symmetric about the y-axis), odd (rotationally symmetric about the  origin) or neither

4th a) find horizontal asymptotes; if the numerator is an nth degree polynomial and the denominator is an mth  degree polynomial, with leading coefficients A in the numerator and B in the denominator, then:

Case 1: m>n  ->  horizontal asymptote at y = 0

Case 2: m=n  ->  horizontal asymptote at y = A/B

Case 3: m  no horizontal asymptote*

b) find vertical asymptotes; if the denominator reaches zero at a value of x, there is a vertical asymptote there.

c) if in Case 3, n=m+1, there will be a diagonal asymptote with slope A/B. Not all slant asymptotes go through the origin; it will be necessary to use long division to find exactly what the asymptote's equation is. (Have questions on how? Leave a comment.)

5th Find f prime, find where it is equal to zero, where it is positive, and where it is negative.

6th Find extrema, by finding critical numbers, places where f prime equals zero or does not exist, and then  checking either the f prime values between critical numbers or the f double prime values at the critical numbers.  The second method will not always work; if f(x)=x^4 , then f'(x)=4x^3 and f''(x)=12x^2 . In this case, the origin  is an extreme, but it is also a critical number when looking for inflection points; the second derivative does not  show whether the curve is concave up or down at that point.

7th Find possible inflection points, values of x where f''=0, and find the concavity between those points.

8th Sketch the curve

Some rules of thumb to keep in mind: find the zeroes of the function and both derivatives; find asymptotes

What exactly is an inflection point?

(An example of the types of questions you could ask here)

If you have an equation such as f(x)=x^4 , the double derivative will be f^(2)(x) = 12x^2 .

This double derivative will equal zero at x = 0 , but f(x) will not change concavity.

An inflection point is defined as a point where the curve changes concavity in the book, but then it is defined as that before double derivatives are brought into the picture; I think it's possible that an inflection point might be defined as simply a point where f''=0 . But then it might not be.

I will try to ask this in class tomorrow, but if I don't, feel free to post your opinion as a comment.