Whenever you're working with a function  with an algebraic definition like the one you're given, your job is to plug in the value (in this case ) wherever  appears in that definition.  This means that your work should look like:

The key here becomes keeping the negative/positive signs in place given all the addition/subtraction and multiplication of negatives in the numerator. Once you've applied the exponents and multiplication in the numerator, you should have:

This then simplifies to , in which the negatives in numerator and denominator cancel, leaving you with just  as your answer.

When you're given a function definition in the form ... as you are here, your job is then to plug in the value in parentheses anywhere that  appears. That means that to solve for  you'll just plug in  for the  in :

And to solve for  you would do the same thing, plugging in  in place of 

To finish the problem, you'll then add 7 + 9 to get the correct answer, 16.

This problem adds a twist to the classic function setup. You should know that whenever you're dealing with a function defined as , your job is to plug in the given input value anywhere there's an  term. Often that value is a number, but here the next thing you're told is that , meaning that your input value is . 

The steps remain the same, however: just plug in  to  and you can solve:

This means that , and you know that it should expand to . So you can set up an equation and solve:

Square the parenthetical term to get:

And then distribute the 3:

And you can now subtract  and  from each side to simplify:

When you divide both sides by 3 the answer should start to become clear:

You cannot get rid of the  term so you can see that  and .  So  satisfies both terms, making  the correct answer.

Whenever you're working with a function defined as , your job is to take the input value--the value in parentheses--and insert it wherever there's an  in the definition. Since here  tells you that your input value is , you can plug that into the function:

You're told that  so you can set up an equation:

And then you can perform the exponent:

And work to get like terms together:

So , meaning that .

The question gives you that  and asks for the value of . The easiest thing to start here is to find the value for  where .

You're given that . You can then set this equal to 0 to get . This means that  has to equal either  or .  In order for  to equal either of these,  must be  or . Only  is a provided answer choice, so  is the correct answer.

When you're working with "nested functions" - problems in which you're asked to apply a function to a function, such as  here - you should follow classic Order-of-Operations and start with the interior parentheses first. Here that means taking , the inner function, and then using that result as your input for the outer function.

In the definition , the left-hand side of the equation defines your "input" saying "whatever you see in the parentheses where  currently is, do to that value what is done to  on the right-hand side of the equation." And then the right-hand side of the equation tells you what to do to your input. Here it's saying "take your input and square it, then subtract one."

When you apply that to , you'll do exactly what that says: plug  into the  spots, meaning you'll take . The result of that is . So .

Now with  your input is , so you'll plug in  for  in . This means that you'll have:

That simplifies to .

This problem tests your familiarity and comfort with function notation. When you're given a definition like , it's important to recognize that  is the "input" (whatever they tell you  is, you then put that into the equation), and that  is the "output" (once you've put your input through the equation, the result is .

Here you're given the function definition and then told that the output, , is equal to . Then you're asked to solve for , which means that you're asked to solve for the input.

So what they're really asking is "for what number, when you take  to that power and then subtract the result from , would you end up with ?"

In equation form, that's . Performing the algebra, you can add  to each side and subtract  from each side.

That gives you: . Knowing your powers of , you should recognize that  and , so  must be between  and .

To solve, you can simply set the output of the function  equal to the algebraic expression: 

And cross-multiply:

And solve for :

Alternatively, you could have recognized that we have a positive result, so the numerator of our fraction cannot be , and that the values for x greater than 1 would all leave a negative denominator, so that cannot be, either.

If , you can simply set the two items equal: . From there, you can "complete the square" by subtracting the right-hand side and moving it to the left, getting to: . This should look familiar as a common algebraic equation; it factors to . Accordingly, the solution must be .

As with many functions problems, this problem is testing whether you can follow directions. You are given two functions and then asked for the value of a nested function, . Remember when you are evaluating functions, that you must always start from the innermost function and work outward since you must always follow the order of operations. So, in this case, the problem is asking you to find the value of  and then to plug the resulting value into .

Plugging in for  is simple: you simply take whatever is the parentheses (in this case, ) and plug it in for  in the function. Your function then transforms from  to .

Your next step will be similar: you just need to take what is in parentheses  and plug that value into the function  every place you see an . This means that  becomes . notice that the entire value  is plugged into the function, not just . If you expand this function using your knowledge of perfect squares, it becomes: , which in turn becomes .

Notice that you could also plug in a number for . If you plug in , the  becomes . As above, remember to start from the inside and work out.  becomes . 

The function  can then be rewritten as . If you then plug in  to the function , you get: .

In order to finish this problem, you simply need to plug your original value for ,  into each answer choice, and look for .