Posted on

## Using This Blog

My name is Stacey and I live in Corvallis, Oregon with my two kids.

This site is created by content from you! I help people on Twitter and online who are stuck on math.  You send me a problem, and I write up a solution and an explanation. Pretty simple.

Almost everything here is free.  Read examples and solutions, browse concepts, learn math.  But, I need to make a living too!

Buy tools I create!  I create practice problems (with solutions), note-cards to use on exams, practice tests and other math-helpers.  These are easy to purchase as you are browsing and practicing.  Don’t see a subject you want to learn about?  Contact me and I’ll create it.

Want your question to appear on my blog? Tweet it to me!

Posted on

## When to use Disk Method versus Shell Method, Part 2

To start, read: When to use Disk Method versus Shell Method, Part 1 to get a general visual.  This is a little more detailed:

Volume should be thought of as infinitely stacked area. In the disk method, you are infinitely stacking circles (think pancakes). In the shell method, you are infinitely stacking lateral surface areas of cylinders (think “Russian Dolls” that stack inside of each other)

The Disk Method: Since you are stacking pancakes, the general formula that you will be integrating is π*r2. If the radius of each disk is changing throughout the shape, the radius, r, will be a function, dependent on either y or x, depending on how you are rotating.

The Shell Method: Since you are stacking lateral areas of cylinders, the general formula that you will be integrating is 2*π*r*h (lateral area of cylinder formula). The radius, r, will be a simple function involving an x or a y. The height, h, will depend on the functions that are being rotated.

Now that you have an understanding of the concept, view this example… I solve a volume problem using the disk method.  Then, I solve the exact same problem using the shell method:   Integration Example: Disk Method vs. Shell Method

Posted on

## Solving logarithmic equations

Q:  Solve
log16x + log4x + log2x = 7

A:  It is easier (though not always necessary) to have all of the logs in the same base to proceed.  Since that is not the case, we need to use the change of base formula to put all of the logs into the same base.

Change of base formula says:
logab = logcb / logcwhere c can be any base of your choosing

What base should we go to?  Since 16, 4 and 2 can all be formed by powers of 2, let’s go to base 2:

log16x + log4x + log2x = 7

log2x / log216 + log2x / log24 + log2x / log22 = 7

Now, we can simplify:

log2x / 4 + log2x / 2 + log2x / 1 = 7

1/4 log2x + 1/2 log2x + log2x = 7

7/4 log2x = 7

log2x = 7 *(4/7)

log2x = 4

(remember, this reads:  the power you put on 2 to get x is 4)

x = 24

x = 16

Posted on

## Solving Exponential Equations by Factoring

Q:  Solve for x:
22x + 2x+2 – 12 = 0

A:  The first thing we need to notice (from practice and experience) is that we can re-write this like so using our knowledge / rules of exponents:

(2x)2 + 22*2x – 12 = 0
Now simplify a little:

(2x)2 + 4*2x – 12 = 0

So, look at this in a new light.  What if we substitute each 2x with y?

(2x)2 + 4*2x – 12 = 0   turns into   y2 + 4*y – 12 = 0

(this isn’t necessary, but is helpful for visualization)

We see this is in the form of a quadratic and can be factored:

(y + 6)(y – 2) = 0

So, y = -6 or y = 2

Remember, y was a substitution for 2x. So, we really have:
2x = -6   or   2x = 2

Posted on

## Solving Percent Problems using Proportions

Q:  How do you solve percent problems using proportions?

A:  Generally, there are 3 main parts to a percent problem:  the percent, the part and the total.

The set up is the same for all problems:

part/total = percent/100

(remember, percent means per 100, so it is always out of 100)

Example 1:  Finding the percent:

What percent is 75 of 85?

Posted on

## Graphing Conic Sections

Q:  Identify the center, foci, vertices, and co vertices then graph the following conic section:
x2 + 9y2 = 36

A:  Look at our given equation:  x2 + 9y2 = 36 and recognize by its form that it is an ellipse.

To match the standard form of an ellipse, we need it equal, so divide both sides by 36 to get:

x2/36 + 9y2/36 = 36/36

Posted on

## Solve the system of equations (conics)

Q:  Solve the system of equations (both of which are conic sections)
1:   x2 + y2 – 20x + 8y + 7 = 0
2:  9x2 + y2 + 4x + 8y + 7 = 0

A:  I am going to solve this by using the elimination method (since I see that I can cancel out all of the y’s)

I am going to multiply equation 1 by (-1) and then add it to equation 2:

2:  9x2 + y2 + 4x + 8y + 7 = 0

1: -x2 – y2 + 20x – 8y – 7 = 0   [after it has been multiplied by -1]

Now, add them together to get equation 3:

3:  8x2 +24x = 0

Posted on

## Finding the limit example

Q: Find the limit (as x approaches 3) of (x3 – 27) / (3-x)

A:  The first thing to do when finding a basic limit is try plugging in the number in question (3).

So, plug in 3 to get:

(33 – 27) / (3-3) = 0/0 <– if you get 0/0 or infinity/infinity that means there is more work to be done.  However, if you had just got a number like 4 or something, that would’ve been your answer!

OK, we got 0/0 so that means more work.  More work could mean many things (apply different rules, factor and cancel, simplify, etc).  In this case, it appears we can factor, so we try that:

(x3 – 27) / (3-x) = (x – 3)(x2 + 3x + 9)/ (3 – x)

Now, here comes some tricky insight.  I notice that the (x – 3) on top is very similar to the (3 – x) on the bottom.  I am going to factor a “-1” out of the (x – 3) that is on top.

Notice:  -1(3 – x) = (x – 3)

So, the numerator becomes:

-1(x – 3)(x2 + 3x + 9)/ (3 – x)

Now, the (x – 3) term cancels from the top and bottom to leave:

-1(x2 + 3x + 9)

So, we are trying to find the limit (as x approaches 3) of -1(x2 + 3x + 9).  We have “removed the hole” — the factor (3 – x) was a “removable hole” that was causing calculation problems.  In the simplified version, we can plug in the value 3 to now calculate where that hole was occuring:

Lim (x –> 3) of -1(x2 + 3x + 9) = -1(32 + 3(3) + 9) = -1(9 + 9 + 9) = -27

Posted on

## Negative Exponent Examples and Basics

Q:  What is a negative exponent?  What does a negative exponent do?

A:  A negative exponent is just notation that can be used to represent a reciprocal.  Let me show you some examples:

x – 1 = 1/x1 or 1/x

x-2 = 1/x2

4 – 1 = 1/41 or 1/4

4 – 2 = 1/42 = 1/16

Those are the basics.  Let’s look at a few more complicated examples of how negative exponents affect an expression:

3x – 4 = 3/x4

4x6y – 3 = 4x6/y3

OR… If the negative exponent is affecting something in a denominator, it will move it to the numerator like so:

1/x – 5 = 1*x5 = x5

6/y – 7 = 6y7

These are just a few basic examples to show you how negative exponents work. Of course things can get more difficult.

Posted on

## Quotient Rule Example

Q:  Find dy/dx of

y = x / sqrt(x2 + 1)

A:  To find dy/dx (the derivative), we will need to use the quotient rule since we have a function over a function.  See? We are in the form:

y = f / g where f and g are two different functions of x.

In this form, the quotient rule tells us if:

y = f / g
then

y ‘ = (g * f ‘ – f * g ‘ ) / g2

So, we know:

f = x

g = sqrt(x2 + 1)

f ‘ = 1   <— basic derivative

g ‘ = (this takes more work…. First, rewrite “g”):

g = sqrt(x2 + 1) = (x2 + 1)1/2

Now, use a power rule and a chain rule to find g ‘ like so:

g ‘ = 1/2 (x2 + 1) – 1/2 (2x)

[that’s the derivative of the outside * the derivative of the inside]

Clean up g ‘:

g ‘ = x (x2 + 1) – 1/2

So now we have all of the players:  f, f ‘ , g, g ‘:

f = x

g = (x2 + 1)1/2

f ‘ = 1

g ‘ = x (x2 + 1) – 1/2

Now, we said:  y ‘ = (g * f ‘ – f * g ‘ ) / g2, so plug in the pieces then simplify like so:

y ‘ = [ (x2 + 1)1/2(1) – (x)(x (x2 + 1) – 1/2 ) ] / [(x2 + 1)1/2]2

CLEAN UP

y ‘ = [ (x2 + 1)1/2 – x2(x2 + 1) – 1/2  ] / (x2 + 1)

Multiply top and bottom by (x2 + 1)1/2

[ (x2 + 1)1/2(x2 + 1)1/2 (x2 + 1)1/2x2(x2 + 1) – 1/2  ] / (x2 + 1)1/2(x2 + 1) =

[ (x2 + 1)- x2 ] / (x2 + 1)3/2

1 / (x2 + 1)3/2

So, if:

y = x / sqrt(x2 + 1)

dy/dx = 1 / (x2 + 1)3/2

Posted on

Q:  Solve for x:  (x + 3)(x – 4) < 0

A:

Step 1:  Find the zeroes

Since the quadratic is already factored, this isn’t too tricky.  If it wasn’t factored, you’d have to factor first! (always make sure there is a 0 on one side of the equation / inequality before proceeding).

OK, so what are the zeroes?

x + 3 = 0 or x – 4 = 0

The zeroes are x = -3 or x = 4

So, draw a number line and plot the zeroes on the number line:

Now, you have to test each “interval” that is separated by the “zeroes”.  There are three intervals to test.

Interval 1:  The numbers to the left of -3  –> in interval notation this is (-infinity, -3)

Interval 2:  The numbers between -3 and 4  –> in interval notation this is (-3, 4)

Interval 3:  The numbers to the right of 4  –> in interval notation this is (4, infinity)

Step 2:  Test each interval

Pick any number on interval 1 and test it into the original inequality.  I’ll pick -5:

(x + 3)(x – 4) < 0

(-5 + 3)(-5 – 4) < 0

(-2)(-9) < 0

18 < 0      <—  this is false, so numbers on this interval [interval 1] are not part of the solution.

Pick any number on interval 2 and test it into the original inequality.  I’ll pick 0:

(x + 3)(x – 4) < 0

(0 + 3)(0 – 4) < 0

(3)(-4) < 0

-12 < 0      <—  this is true, so numbers on this interval [interval 2] are a part of the solution.

Pick any number on interval 3 and test it into the original inequality.  I’ll pick 5:

(x + 3)(x – 4) < 0

(5 + 3)(5 – 4) < 0

(8)(1) < 0

8 < 0      <—  this is false, so numbers on this interval [interval 3] are not part of the solution.

SO:  The only interval that “worked” was interval 2.  Therefore, the solution is all number between -3 and 4.

In interval notation, we write that like: (-3, 4)

In inequality notation, we write that like: -3 < x < 4

Posted on

## Critical points, Max/Min, Points of Inflection, Concavity – all in one!

Q:  Consider the function:  f(x) = (1/3) x3 – x2 – 15x + 1

(a) critical points

(b) the local minimum / maximums [relative extrema]

(c) points of inflection

(d) concavity

A:  OK.  Get out your pen and paper and be writing with me as I calculate and explain.  This is the best way for you to learn.

(a)  The critical points are where the derivative equals 0 or is undefined.  So, take the derivative to start:

f(x) = (1/3) x3 – x2 – 15x + 1

f ‘ (x) = x2 – 2x- 15  <– you should get this!

Now, we need to find where the derivative = 0 (or is undefined — since the derivative is a quadratic, it is never undefined, so we don’t have to worry about that part).

0 = x2 – 2x- 15

Solve by factoring:

0 = (x – 5)(x + 3)

So, the critical points are when x = 5, -3.

(b)  To determine if the critical points are local maximum or minimums, we need to do the second derivative test.

Recap on second derivative test:  Plug a critical point into the second derivative.

*If the end result is positive, this means the shape is concave up (smiley face) which means we have a minimum.

*If the end result is negative, this means the shape is concave down (frowny face) which means we have a maximum.

So, take the second derivative:

f(x) = (1/3) x3 – x2 – 15x + 1  [original function]

f ‘ (x) = x2 – 2x- 15  [1st derivative]

f ” (x) = 2x – 2  [2nd derivative]

Now, we have critical points of x = 5 and x = -3.  Deal with them one at a time in the second derivative to perform the second derivative test.

f ” (5) = 2(5) – 2 = 10 – 2 = 8 <– this is positive, which means when x = 5 we have a local minimum!

f ” (-3) = 2(-3) – 2 = -6 – 2 = -8 <– this is negative, which means when x = -3 we have a local maximum!

If you wanted to find the (x, y) point of the actual max and min, you would plug the x-values into the original equations to find the corresponding y values:

f(5) = (1/3) (5)3 – (5)2 – 15(5) + 1 = -172/3

So, f(x) has a local minimum at the point (5, -172/3)

f(-3) = (1/3) (-3)3 – (-3)2 – 15(-3) + 1 = 28

So, f(x) has a local maximum at the point (-3, 28)

(c)  The points of inflection are where the concavity of the function changes from concave up to concave down (and vice versa).  Our POI candidates can be found by setting the second derivative = 0 and finding the x candidates:

f ” (x) = 2x – 2

0 = 2x – 2 and solve for x:

x = 1

So, x = 1 is a candidate for a point of inflection (POI) — it is not a guaranteed POI though.

We need to interval test into the second derivative to see what is happening (generally, you would need to test in between every x value.  We only have 1 x value, so we just need to interval test on either side of it):

Plot your x values on a number line.  In our case, the only value is x = 1.  Pick a number on the left side of 1.  Let’s pick an easy number, 0:

Plug 0 into the second derivative:

f ” (0) = 2(0) – 2 = -2 <– since this is negative:  this interval, to the left of x = 1, is concave down

Now, pick a number to the right of x = 1.  Let’s pick 2:

f ” (2) = 2(2) – 2 = 2 <– since this is positive:  this interval, to the right of x = 1, is concave up

Therefore, when x = 1 we do have a point of inflection (since we changed from concave down to concave up).

If you want the (x, y) point, plug the x value into the original equation to find y:

f(1) = (1/3) (1)3 – (1)2 – 15(1) + 1 = -44/3

There is a point of inflection at (1, -44/3).

(d)  Now to analyze concavity, which we have basically already done in part (c)!

To the left of x = 1, we determined the graph to be concave down.

To the right of x = 1, we determined the graph to be concave up.

Concave down on the interval (-infinity, 1)

Concave up on the interval (1, infinity)
We have done it.  That was a lot to do.  Enjoy.