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1. This is asking me for all the values of (g o f)(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.
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Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. The.NET Framework 2.0 Service Pack 2 provides cumulative roll-up updates for customer reported issues found after the release of the.NET Framework 2.0. Backyard baseball 2. In addition, this release provides performance improvements, and prerequisite feature support for the.NET Framework 3.5 Service Pack 1. In mathematics, trigonometric substitution is the substitution of trigonometric functions for other expressions. In calculus, trigonometric substitution is a technique for evaluating integrals. Moreover, one may use the trigonometric identities to simplify certain integrals containing radical expressions. Like other methods of integration by substitution, when evaluating a definite integral.

Unlike the software developed for Windows system, most of the applications installed in Mac OS X generally can be removed with relative ease. goPanel 1.0.3 is a third party application that provides additional functionality to OS X system and enjoys a popularity among Mac users. However, instead of installing it by dragging its icon to the Application folder, uninstalling goPanel 1.0.3 may need you to do more than a simple drag-and-drop to the Trash.

When installed, goPanel 1.0.3 creates files in several locations. Generally, its additional files, such as preference files and application support files, still remains on the hard drive after you delete goPanel 1.0.3 from the Application folder, in case that the next time you decide to reinstall it, the settings of this program still be kept. But if you are trying to uninstall goPanel 1.0.3 in full and free up your disk space, removing all its components is highly necessary. Continue reading this article to learn about the proper methods for uninstalling goPanel 1.0.3.

Manually uninstall goPanel 1.0.3 step by step:

Most applications in Mac OS X are bundles that contain all, or at least most, of the files needed to run the application, that is to say, they are self-contained. Thus, different from the program uninstall method of using the control panel in Windows, Mac users can easily drag any unwanted application to the Trash and then the removal process is started. Despite that, you should also be aware that removing an unbundled application by moving it into the Trash leave behind some of its components on your Mac. To fully get rid of goPanel 1.0.3 from your Mac, you can manually follow these steps:

Before uninstalling goPanel 1.0.3, you'd better quit this application and end all its processes. If goPanel 1.0.3 is frozen, you can press Cmd +Opt + Esc, select goPanel 1.0.3 in the pop-up windows and click Force Quit to quit this program (this shortcut for force quit works for the application that appears but not for its hidden processes).

Open Activity Monitor in the Utilities folder in Launchpad, and select All Processes on the drop-down menu at the top of the window. Select the process(es) associated with goPanel 1.0.3 in the list, click Quit Process icon in the left corner of the window, and click Quit in the pop-up dialog box (if that doesn't work, then try Force Quit).

First of all, make sure to log into your Mac with an administrator account, or you will be asked for a password when you try to delete something.

Open the Applications folder in the Finder (if it doesn't appear in the sidebar, go to the Menu Bar, open the 'Go' menu, and select Applications in the list), search for goPanel 1.0.3 application by typing its name in the search field, and then drag it to the Trash (in the dock) to begin the uninstall process. Alternatively you can also click on the goPanel 1.0.3 icon/folder and move it to the Trash by pressing Cmd + Del or choosing the File and Move to Trash commands.

For the applications that are installed from the App Store, you can simply go to the Launchpad, search for the application, click and hold its icon with your mouse button (or hold down the Option key), then the icon will wiggle and show the 'X' in its left upper corner. Click the 'X' and click Delete in the confirmation dialog.

Though goPanel 1.0.3 has been deleted to the Trash, its lingering files, logs, caches and other miscellaneous contents may stay on the hard disk. For complete removal of goPanel 1.0.3, you can manually detect and clean out all components associated with this application. You can search for the relevant names using Spotlight. Those preference files of goPanel 1.0.3 can be found in the Preferences folder within your user's library folder (~/Library/Preferences) or the system-wide Library located at the root of the system volume (/Library/Preferences/), while the support files are located in '~/Library/Application Support/' or '/Library/Application Support/'.

Open the Finder, go to the Menu Bar, open the 'Go' menu, select the entry:|Go to Folder. and then enter the path of the Application Support folder:~/Library

Search for any files or folders with the program's name or developer's name in the ~/Library/Preferences/, ~/Library/Application Support/ and ~/Library/Caches/ folders. Right click on those items and click Move to Trash to delete them.

Meanwhile, search for the following locations to delete associated items:

• /Library/Preferences/
• /Library/Application Support/
• /Library/Caches/

Besides, there may be some kernel extensions or hidden files that are not obvious to find. In that case, you can do a Google search about the components for goPanel 1.0.3. Usually kernel extensions are located in in /System/Library/Extensions and end with the extension .kext, while hidden files are mostly located in your home folder. You can use Terminal (inside Applications/Utilities) to list the contents of the directory in question and delete the offending item.

If you are determined to delete goPanel 1.0.3 permanently, the last thing you need to do is emptying the Trash. To completely empty your trash can, you can right click on the Trash in the dock and choose Empty Trash, or simply choose Empty Trash under the Finder menu (Notice: you can not undo this act, so make sure that you haven't mistakenly deleted anything before doing this act. If you change your mind, before emptying the Trash, you can right click on the items in the Trash and choose Put Back in the list). In case you cannot empty the Trash, reboot your Mac.

Tips for the app with default uninstall utility:

You may not notice that, there are a few of Mac applications that come with dedicated uninstallation programs. Though the method mentioned above can solve the most app uninstall problems, you can still go for its installation disk or the application folder or package to check if the app has its own uninstaller first. If so, just run such an app and follow the prompts to uninstall properly. After that, search for related files to make sure if the app and its additional files are fully deleted from your Mac.

Automatically uninstall goPanel 1.0.3 with MacRemover (recommended):

No doubt that uninstalling programs in Mac system has been much simpler than in Windows system. But it still may seem a little tedious and time-consuming for those OS X beginners to manually remove goPanel 1.0.3 and totally clean out all its remnants. Why not try an easier and faster way to thoroughly remove it?

If you intend to save your time and energy in uninstalling goPanel 1.0.3, or you encounter some specific problems in deleting it to the Trash, or even you are not sure which files or folders belong to goPanel 1.0.3, you can turn to a professional third-party uninstaller to resolve troubles. Here MacRemover is recommended for you to accomplish goPanel 1.0.3 uninstall within three simple steps. MacRemover is a lite but powerful uninstaller utility that helps you thoroughly remove unwanted, corrupted or incompatible apps from your Mac. Now let's see how it works to complete goPanel 1.0.3 removal task.

The whole uninstall process may takes even less than one minute to finish, and then all items associated with goPanel 1.0.3 has been successfully removed from your Mac!

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Benefits of using MacRemover:

MacRemover has a friendly and simply interface and even the first-time users can easily operate any unwanted program uninstallation. With its unique Smart Analytic System, MacRemover is capable of quickly locating every associated components of goPanel 1.0.3 and safely deleting them within a few clicks. Thoroughly uninstalling goPanel 1.0.3 from your mac with MacRemover becomes incredibly straightforward and speedy, right? You don't need to check the Library or manually remove its additional files. Actually, all you need to do is a select-and-delete move. As MacRemover comes in handy to all those who want to get rid of any unwanted programs without any hassle, you're welcome to download it and enjoy the excellent user experience right now!

This article provides you two methods (both manually and automatically) to properly and quickly uninstall goPanel 1.0.3, and either of them works for most of the apps on your Mac. If you confront any difficulty in uninstalling any unwanted application/software, don't hesitate to apply this automatic tool and resolve your troubles.

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).

Double down casino slots free. (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:

General purpose cad software. (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
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(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.

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.

• 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.

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).

Double down casino slots free. (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:

General purpose cad software. (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
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(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.

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.

• 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.
  

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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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