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PicGIF for Mac is an easy to use and intuitive GIF maker that allows you to create animated GIFs from your life photos and videos on Mac OS X. Free to try PearlMountain Technology Mac OS X 10. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history. Last Updated on September 9, 2016 by admin. Description. Limited Time Offer – 70% OFF! Back to $19.90 Soon. Convert photos and videos into incredible GIFs quickly.

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PicGIF for Mac Editor's Review

Create animated GIF's from your life pictures with this handy Mac tool.

Some of the simplest movies that you can create are simply made from stills and stitched together with special tools. I've done this throughout my computing career for a number of different reasons. However, that's mostly been on the Windows side of the house. Thankfully, you can do this just as easily on the Mac side of the world, and that's one of the reasons why I really like PicGIF from PearlMountain Software. It's an animated GIF tool for your Mac, and its really easy to use.

PicGIF for Mac is an easy to use and intuitive GIF maker that allows you to create animated GIFs from your photos and videos on your Mac. One of its really cool features also lets you edit existing animated GIFs. The program is simple and easy to use, and it makes animating your pictures a lot of fun.

One of the coolest things about PicGIF is that it doesn't matter what format your source photos are in. If you're Mac can open them, then Pic GIF can use them to make an animated GIF. The app also works with common video formats, allowing you to quickly and easily string videos together into a longer, larger movie that can be played over the web or by other computers. It's a great way to quickly and easily share special moments with family members and friends who happen to live far away.

App Pro's: Supports multiple still and video formats, easy to use

App Con's: Requires OS X 10.7 Lion or later

Conclusion: PicGIF is a great application. Not only can you take control of the entire animation process, but you can set the picture size, fill mode, frame delay and playing sequence. Because its on your Mac, you can do most of the things that you would expect you can do with pictures on a Mac. It supports full drag and drop, has a text editor that will allow you to add text to your animations as well as giving you complete font control. You also get a real time preview of the animation. The app is amazing.

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Composition of Functions:
Composing with Sets of Points (page 1 of 6)

Sections: Composing functions that are sets of point, Composing functions at points, Composing functions with other functions, Word problems using composition, Inverse functions and composition

Until now, given a function f(x), you would plug a number or another variable in for x. You could even get fancy and plug in an entire expression for x. For example, given f(x) = 2x + 3, you could find f(y² – 1) by plugging y² – 1 in for x to get f(y² – 1) = 2(y² – 1) + 3 = 2y² – 2 + 3 = 2y² + 1.

In function composition, you're plugging entire functions in for the x. In other words, you're always getting 'fancy'. But let's start simple. Instead of dealing with functions as formulas, let's deal with functions as sets of (x, y) points:

• Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and
  let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}.
  Find (i)f (1), (ii) g(–1), and (iii) (gof )(1).

(i) This type of exercise is meant to emphasize that the (x, y) points are really (x, f (x)) points. To find f (1), I need to find the (x, y) point in the set of (x, f (x)) points that has a first coordinate of x = 1. Then f (1) is the y-value of that point. In this case, the point with x = 1 is (1, –1), so:

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f (1) = –1

(ii) The point in the g(x) set of point with x = –1 is the point (–1, –2), so:

g(–1) = –2

(iii) What is '(gof )(1)'? This is read as 'g-compose-f of 1', and means 'plug 1 into f, evaluate, and then plug the result into g'. The computation can feel a lot easier if I use the following, more intuitive, formatting:

(gof )(1) = g( f(1))

Now I'll work in steps, keeping in mind that, while I may be used to doing things from the left to the right (because that's how we read), composition works from the right to the left (or, if you prefer, from the inside out). So I'll start with the x = 1. I am plugging this into f(x), so I look in the set of f(x) points for a point with x = 1. The point is (1, –1). This tells me that f(1) = –1, so now I have: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved


(gof )(1) = g( f(1)) = g(–1)

Working from the right back toward the left, I am now plugging x = –1 (from 'f(1) = –1') into g(x), so I look in the set of g(x) points for a point with x = –1. That point is (–1, –2). This tells me that g(–1) = –2, so now I have my answer:

(gof )(1) = g( f(1)) = g(–1) = –2

Note that they never told us what were the formulas, if any, for f(x) or g(x); we were only given a list of points. But this list was sufficient for answering the question, as long as we keep track of our x- and y-values.

• Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and
  let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}.
  
  Find (i) ( fog)(0), (ii)( fog)(–1), and (iii)(gof )(–1).

(i) To find ( fog)(0), ('f-compose-g of zero'), I'll rewrite the expression as:

( fog)(0) = f(g(0))

This tells me that I'm going to plug zero into g(x), simplify, and then plug the result into f(x). Looking at the list of g(x) points, I find (0, 2), so g(0) = 2, and I need now to find f(2). Looking at the list of f(x) points, I find (2, –3), so f(2) = –3. Then:

( fog)(0) = f(g(0)) = f(2) = –3

(ii) The second part works the same way:

( fog)(–1) = f(g(–1)) = f(–2) = 3

(iii) I can rewrite the composition as (gof )(–1) = g( f(–1)) = g(1).

Uh-oh; there is no g(x) point with x = 1, so it is nonsense to try to find the value of g(1). In math-speak, g(1) is 'not defined'; that is, it is nonsense.Then (gof )(–1) is also nonsense, so the answer is:

(gof )(–1) is undefined.

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Part (iii) of the above example points out an important consideration regarding domains and ranges. It may be that your composed function (the result you get after composing two other functions) will have a restricted domain, or at least a domain that is more restricted than you might otherwise have expected. This will be more important when we deal with composing functions symbolically later.

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Another exercise of this type gives you two graphs, rather than two sets of points, and has you read the points (the function values) from these graphs. Folx pro 5 4 – download manager windows 10.

• Given f(x) and g(x) as shown below, find ( fog)(–1).
  

In this case, I will read the points from the graph. I've been asked to find ( fog)(–1) = f(g(–1)). This means that I first need to find g(–1). So I look on the graph of g(x), and find x = –1. Tracing up from x = –1 to the graph of g(x), I arrive at y = 3. Then the point (–1, 3) is on the graph of g(x), and g(–1) = 3.

Now I plug this value, x = 3, into f(x). To do this, I look at the graph of f(x) and find x = 3. Tracing up from x = 3 to the graph of f(x), I arrive at y = 3. Then the point (3, 3) is on the graph of f(x), and f(3) = 3.

Then( fog)(–1) = f(g(–1)) = f(3) = 3.

• Given f(x) and g(x) as shown in the graphs below, find (gof )(x) for integral values of x on the interval –3 <x< 3.
  

This is asking me for all the values of (gof )(x) = g( f(x)) for x = –3, –2, –1, 0, 1, 2, and 3. So I'll just follow the points on the graphs and compute all the values:

(gof )(–3) = g( f(–3)) = g(1) = –1

I got this answer by looking at x = –3 on the f(x) graph, finding the corresponding y-value of 1 on the f(x) graph, and using this answer as my new x-value on the g(x) graph. That is, I looked at x = –3 on the f(x) graph, found that this led to y = 1, went to x = 1 on the g(x) graph, and found that this led to y = –1. Similarly:

(gof )(–2) = g( f(–2)) = g(–1) = 3
(gof )(–1) = g( f(–1)) = g(–3) = –2
(gof )(0) = g( f(0)) = g(–2) = 0
(gof )(1) = g( f(1)) = g(0) = 2
(gof )(2) = g( f(2)) = g(2) = –3
(gof )(3) = g( f(3)) = g(3) = 1

You aren't generally given functions as sets of points or as graphs, however. Generally, you have formulas for your functions. So let's see what composition looks like in that case..

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Another exercise of this type gives you two graphs, rather than two sets of points, and has you read the points (the function values) from these graphs. Folx pro 5 4 – download manager windows 10.

• Given f(x) and g(x) as shown below, find ( fog)(–1).
  

In this case, I will read the points from the graph. I've been asked to find ( fog)(–1) = f(g(–1)). This means that I first need to find g(–1). So I look on the graph of g(x), and find x = –1. Tracing up from x = –1 to the graph of g(x), I arrive at y = 3. Then the point (–1, 3) is on the graph of g(x), and g(–1) = 3.

Now I plug this value, x = 3, into f(x). To do this, I look at the graph of f(x) and find x = 3. Tracing up from x = 3 to the graph of f(x), I arrive at y = 3. Then the point (3, 3) is on the graph of f(x), and f(3) = 3.

Then( fog)(–1) = f(g(–1)) = f(3) = 3.

• Given f(x) and g(x) as shown in the graphs below, find (gof )(x) for integral values of x on the interval –3 <x< 3.
  

This is asking me for all the values of (gof )(x) = g( f(x)) for x = –3, –2, –1, 0, 1, 2, and 3. So I'll just follow the points on the graphs and compute all the values:

(gof )(–3) = g( f(–3)) = g(1) = –1

I got this answer by looking at x = –3 on the f(x) graph, finding the corresponding y-value of 1 on the f(x) graph, and using this answer as my new x-value on the g(x) graph. That is, I looked at x = –3 on the f(x) graph, found that this led to y = 1, went to x = 1 on the g(x) graph, and found that this led to y = –1. Similarly:

(gof )(–2) = g( f(–2)) = g(–1) = 3
(gof )(–1) = g( f(–1)) = g(–3) = –2
(gof )(0) = g( f(0)) = g(–2) = 0
(gof )(1) = g( f(1)) = g(0) = 2
(gof )(2) = g( f(2)) = g(2) = –3
(gof )(3) = g( f(3)) = g(3) = 1

You aren't generally given functions as sets of points or as graphs, however. Generally, you have formulas for your functions. So let's see what composition looks like in that case..

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