All SAT Mathematics Resources
Example Questions
Example Question #1 : Functions
The function is defined above. What is ?
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When you're given the definition of a function as you are here, , your job to calculate a function is to take the value in parentheses and plug that in for wherever it appears in the definition. Here, qualitatively, you're being told "whatever is, square it and then subtract from that square." That means that:
So:
And for you'd have:
So:
This means that , so the correct answer is .
Example Question #1 : Functions
If , what is ?
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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.
Example Question #1 : Functions
The function is defined above. What is ?
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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.
Example Question #1 : Working With Function Notation
If and , what is the value of ?
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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.
Example Question #2 : Functions
The function is defined for all values as , where is a constant. If , then what is the value of ?
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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 .
Example Question #1 : Functions
The function is defined as . If , what is one possible value for ?
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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.
Example Question #2 : Working With Function Notation
The function is defined for all real numbers as . What is ?
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 .
Example Question #3 : Functions
The function is defined by . If , then which of the following is true about the value of ?
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 .
Example Question #1 : Functions
If for all values of , for what value of does ?
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.
Example Question #1 : Functions
If and , for which of the following values does ?
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 .
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