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Limits vs Values

Look at the function f(x) in orange below:

limit6

We are going to answer 4 questions about this graph.  They are all related to each other, but different questions.  Seeing the difference will help us sort out the difference between a function value and a limit.

Q1:  Find f(1)

Q2:  Find  limx→1 f(x)

Q3:  Find  limx→1+ f(x)

Q4:  Find  limx→1 f(x)

OK….. Try to answer these questions with what you know… Then continue reading to see the answers and explanations!

Continue reading Limits vs Values

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Basic Concept of a Limit

Here is a brief example on the concept of a limit:

Look at the function f(x) in orange below:

limit1

Question 1.  Find f(3)

Explanation of question 1: Find the value of the function when you plug in 3.  What is the height of the function at the exact moment when x=3?

Answer 1:  The function is undefined at x=3.  There is a hole when x=3.

So, f(3) is undefined.

Question 2:  Find limx→3f(x)

Explanation of question 2:  We are being asked to find what the function is doing around (but not at) 3.  What is happening to the path of the function on either side of 3?

In order to find limx→3f(x), we must confirm that limx→3+  f(x) and limx→3–  f(x) both exist and are equal to each other.

So, let’s find limx→3+  f(x).  What is happening to the function values as you approach x=3 from the right-hand side?  Literally run your finger along as if x=4, then x=3.5, then x=3.1.  What value is the function getting closer to?

limit2

The function is approaching a height of 4.

Let’s find limx→3–  f(x).  What is happening to the function values as you approach x=3 from the left-hand side?  Literally run your finger along as if x=1, then x=2, then x=2.9.  What value is the function getting closer to?

limit3

The function is also approaching a height of 4.

So:

limx→3+  f(x) = 4

limx→3–  f(x) = 4

Since, the left-handed limit at 3 and right-handed limit at 3 exist and are equal, this gives:

limx→3f(x) = 4.

So, to summarize, here are 4 different things we found.  They are related, but not necessarily the same:

f(3) is undefined

limx→3+  f(x) = 4

limx→3–  f(x) = 4

limx→3f(x) = 4

Are you ready to try one on your own? Click here! (Don’t worry, I’ll walk you through the solutions too)

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Limit Example

Q:  What is limx→2 (x-2)/|x-2|

A:  This question is not too bad if you know what the function (x-2)/|x-2| looks like graphically.  But, let’s say you don’t.

We are going to “talk our way” through this problem to help solidify the concept of a limit.

If you plug in 2 to the function, you are finding the value of the function when x=2.  This is important, and related, though it is not the limit.  This is even sometimes a skill used to help us find the limit, but it is still not the limit.  Sometimes the function value is equal to the function limit, which can also be confusing, but not all the time.  Let’s find the value of the function when x=2:

(2-2)/|2-2| = 0/0 = undefined.

Okay.  This function is undefined when x=2.  This means there is a hole, or an asymptote, or a break or a jump or some disruption in the continuity of the function. Continue reading Limit Example

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One-sided Limit Example

Q:  Find the one-sided limit (if it exists):

limx→-1–     (x+1)/(x4-1)

A:  So we need to find the limit of this function (x+1)/(x4-1) as x approaches -1 from the left.  Remember, from the left means as x gets closer and closer to -1, but is still smaller.

The concept: What is happening to this function as x = -2, x = -1.5, x = -1.1, x = -1.0001, etc…

We test first and plug -1 into the function: (-1+1)/((-1)4-1) = 0/0

Whenever you get 0/0, that is your clue that maybe you need to do “more work” before just plugging in or jumping to conclusions.

So, let’s try “more work” — usually that means simplifying.  I see that the denominator can factor.  We have:

(x+1) / (x4-1) = (x+1) / [(x2-1)(x2+1)]

Let’s keep factoring the denominator:

(x+1) / [(x-1)(x+1)(x2+1)]

Now, it appears there is a “removable hole” in the function.  This means, we can remove this hole by reducing the matching term in the numerator with the matching term in the denominator:

(x+1) / [(x-1)(x+1)(x2+1)]

= 1 / [(x-1)(x2+1)]

Notice that hole exists when x = -1 (and it was removable! This is good news for us since we are concerned with the nature of the function as x approaches -1)

Continue reading One-sided Limit Example

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Graphing Lines by Plotting Points

Q:  Graph x + 2y = 4

A:  We are going to graph this by making an x-y table of values.  A typical (though not necessary) thing to do is solve for y first — this just means “get y by itself”

So, start with the equation:

x + 2y = 4

Subtract x from both sides (to try to get y alone)

2y = 4 – x

Now, it’s standard for to list the x first, so we may write this as:

2y = -x + 4

To get y by itself, divide everything by 2:

y = -x/2 + 4/2

This simplifies to:

y = -1/2 x + 2

So, to make a table of values, pick some numbers for x.  It is standard to pick “0, 1, 2” or something similar.  So:

x  |  0  |  1  |  2  |

y  |      |      |      |

Plug in each value for x to find y:

When x = 0, y = -1/2 (0) + 2 = 0 + 2 = 2

When x = 1,  y = -1/2 (1) + 2 = -1/2 + 2 = 1.5

When x = 2, y = -1/2 (2) + 2 = -1 + 2 = 1

Then the table is complete:

x  |  0  |  1    |  2  |

y  |  2  | 1.5  |  1  |

These are just points you need to plot on a graph:  (0, 2)  (1, 1.5)  and (2,1)

Plot the points then connect the dots to make a line!

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Graphing Quadratics by creating a table

Q:  Sketch graphs of these three quadratics relations on the same set of axes

a) y=-3x2

b) y=1/4 x2
A:  There are many ways to sketch graphs (quadratics).  The most basic way is to:  (1) make a table of values, (2) plot the points, (3) connect the dots.

So, let’s start with the first equation (a) y=-3x2 and make a table of values:

STEP 1

Start by picking values for x.  I will pick -2, -1, 0, 1, 2 for x:

x  | -2 | -1 | 0 | 1 | 2

y  |     |     |      |    |

Now we need to calculate the y values.

When x = -2, y=-3(-2)2 = -3(4) = -12

When x = -1, y=-3(-1)2 = -3(1) = -3

When x = 0, y=-3(0)2 = -3(0) = 0

When x = 1, y=-3(1)2 = -3(1) = -3

When x = 2, y=-3(2)2 = -3(4) = -12

So, the table is complete:

x  | -2   |  -1  | 0   | 1   | 2

y  | -12 |  -3 |  0  | -3 |  -12

STEP 2:

Plot the points: (-2, -12)  (-1, -3)  (0, 0)  (1, -3)  (2, -12)  on the axes

STEP 3

Connect the dots to make a parabola.  You should have a picture like so:

eq1

Now we need to do the same thing for equation (b) y=1/4 x2

STEP 1

Start by picking values for x.  I will pick -2, -1, 0, 1, 2 for x:

x  | -2 | -1 | 0 | 1 | 2

y  |     |     |      |    |

Now we need to calculate the y values.

When x = -2, y=1/4 (-2)2 = 1/4 (4) = 1

When x = -1, y=1/4 (-1)2 = 1/4 (1) = 1/4

When x = 0, y=1/4 (0)2 = 1/4 (0) = 0

When x = 1, y=1/4 (1)2 = 1/4 (1) = 1/4

When x = 2, y=1/4 (2)2 = 1/4 (4) = 1

So, the table is complete:

x  | -2   |  -1    | 0   |   1   | 2

y  |  1    |  1/4 |  0  | 1/4 |  1

STEP 2:

Plot the points: (-2, 1)  (-1, 1/4)  (0, 0)  (1, 1/4)  (2, 1)  on the axes

STEP 3

Connect the dots to make a parabola.  You should have a picture like so:

eq2

SO…. If we wanted to put these two graphs on the same axes, it would look something like (depending on your scale on the y-axis):

eq3

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Amplitude, Phase Shift and Period

Q:  Consider the function y = -5 cos (2x – .1π)

a)  Identify the amplitude

b)  Identify the phase shift

c)  Identify the period

Answer:

The following formulas / concepts will work for sin and cos graphs:

y = a cos (b*(x – h) ) + k

Note:  a, b, h and k are just numbers that will affect the graph.

The amplitude is |a|

The phase shift if 2π / |b|

The phase shift is “h”

The vertical shift is “k”

A big thing to notice:  the “b” value is factored out in this red formula.

So…. Let’s start with our actual example: y = -5 cos (2x – .1π)

Identify who is a, b, h, and k.  Notice, the b is not factored out, so let’s do that firt:

y = -5 cos (2x – .1π)

y = -5 cos (2*(x – .05π))

[if you multiply the 2 back through, you get the same as the original]

Now,

a = -5

b = 2

h = .05*pi

k = 0 (there is no + k at the end of the problem)

a)  The amplitude is |a| = |-5| = 5

b)  The phase shift is h = .05π (the graph is shifted .05π units to the right)

c)  The period is 2π/|b| = 2π/|2| = π

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Graphing a Quadratic Function: 2 methods

Q:  Graph the Quadratic Function:  f(x) = 2x2 – 3x + 1

Answer:

There are two main methods to do this.  I will do both methods (depending on what you have learned in class, you will want to pick either method 1 or method 2 that I show).

Method 1:  Plotting Points

Because this is a quadratic equation, we know the graph will be in the shape of a parabola.  So, I am going to pick a few values for x, then find the y values.  Then, I will plot the points on the graph to make a parabola.  You can pick any values for x when you do this method.

Let’s pick x = 0.  So, plug x into the equation to find y:

2(0)2 – 3(0) + 1 = 0 – 0 + 1 = 1

Point 1:  When x = 0, y = 1:  (0, 1)


Let’s pick x = 1 now:

2(1)2 – 3(1) + 1= 2(1) – 3 + 1 = 2 – 3 + 1 = 0

Point 2:  When x = 1, y = 0:  (1, 0)


Let’s pick x = -1 now:

2(-1)2 – 3(-1) + 1= 2(1) + 3 + 1 = 2 + 3 + 1 = 6

Point 3:  When x = -1, y = 6:  (-1, 6)

Now, plot the points (0, 1) and (1, 0) and (-1, 6) on a graph.  Connect the dots to make a parabola!

Keep in mind:  You may need to get more than 3 points for better accuracy.

Method 2:  Finding the y-intercept and the vertex:

The y-intercept of any type of graph occurs when you plug in 0 for x.  So, plug in 0 for x into the equation and solve for y:

Continue reading Graphing a Quadratic Function: 2 methods

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Quadratic Example (all parts)

Q:  Consider the equation:

y= x²-4x+3

1. Find the vertex.
2. It opens _____.
3. Find the axis of symmetry.
4. Find the minimum.
5. Find the discriminant.
6. It has ___ real solutions.
7. Find the x-intercept(s), if none, say so.
8. Find the y-intercept.

Continue reading Quadratic Example (all parts)