All SAT Math Resources
Example Questions
Example Question #23 : How To Find F(X)
If z + 2x = 10 and 7z + 2x = 16, what is z?
-2
1
2
-1
0
1
Subtract the first expression from the second. That gives you 6z = 6. That simplifies to z = 1.
Example Question #2771 : Sat Mathematics
In the following sequence:
8, 11, 14, 17, 20...
Which of the following equations could represent the underlying function of this sequence?
3n + 4
3n + 5
2n + 4
2n + 5
4n + 4
3n + 5
In this question, the first 5 terms of the sequence are given. In other words, when the inputs n = {1, 2, 3, 4, 5}, an underlying function f(n) produces an output of {8, 11, 14, 17, 20}.
The best way to solve this problem is to use the answers.
The answers have 5 possible functions for this sequence.
2n + 4
2n + 5
3n + 4
3n + 5
4n + 4
When n = {1, 2, 3, 4, 5}, these functions produce the following sequence:
2n + 4 = {6, 8, 10, 12, 14}
2n + 5 = {7, 9, 11, 13, 15}
3n + 4 = {7, 10, 13, 16, 19}
3n + 5 = {8, 11, 14, 17, 20}
4n + 4 = {8, 12, 16, 20, 24}
Therefore the function 3n + 5 is the correct answer.
Note: when determining if a certain function is the correct one, you do not have to solve for all 5 terms. As soon as one of the terms do not match the original sequence, that function can be eliminated.
Example Question #2772 : Sat Mathematics
Let f(x) be a function with at least one root. All of the following graphs must have the exact same root(s) as f(x) EXCEPT:
(f(x))–1
–4f(x)
(f(x))(1/2)
(f(x))2
|f(x)|
(f(x))–1
Remember that a root of a function is the point where it crosses the x-axis, i.e. an x-intercept. The x-intercepts of a function occur where the y-value of a point is equal to zero. Therefore, the solutions to the equation f(x) = 0 give us the roots of f(x). Thus, if we set the functions in the answer choices equal to 0, and we end up solving the equation f(x) = 0, then the function will have the same roots as f(x) .
Let's set each of the functions in the answer choices equal 0.
First, let's look at (f(x))(1/2) = 0. Raising a function to the 1/2 power is the same as taking the square root of f(x). To get rid of the square root, we can square both sides.
(f(x))(1/2) = √(f(x)) = 0
(√f(x))2 = 02
f(x) = 0. Solving the equation f(x) = 0 will give us the roots of f(x). This means, to find the roots of (f(x))(1/2), we will end up having to find the roots of f(x). In short, the roots of (f(x))(1/2) will be the same as those of f(x).
Intuitively, this makes sense. When we graph (f(x))(1/2) , what we are doing is taking the square root of the y-values of every point on f(x). The roots of f(x) will all have y-values of zero, and taking the square root of zero isn't going to change the y-value. Thus, the location of the roots on f(x) will not change, and the roots of (f(x))(1/2) will be the same as the roots of f(x).
Next, we can look at (f(x))2. We can set it equal to zero to find the roots.
(f(x))2 = 0
We can then take the square root of both sides. The square root of zero is zero.
f(x) = 0
Once again, we are left with the equation f(x) = 0, which will give us the roots of f(x). This makes sense, because (f(x))2 means we are squaring the y-value of the roots on f(x), which wouldn't change the location of the roots, because the square of zero is still zero.
The third graph is –4f(x). Once again, we can set it equal to zero.
–4f(x) = 0
Divide both sides by –4.
f(x) = 0
We are left with this equation, which will give us the roots of f(x). This makes sense, because if we were to multiply the y-values of the x-intercepts on f(x) by –4, then the location of the points wouldn't change. Multiply 0 by –4 will still give us 0.
The next function is |f(x)|. Whenever we take an absolute value of a function, we take any negative points and make them positive. Essentially, |f(x)| reflects all of the negative points on f(x) across the x-axis. However, points located on the x-axis are unchanged by taking the absolute value, because the absolute value of zero is still going to be zero. In other words, if we take the absolute value of the zeros on f(x), they will still be at the same location. Thus, |f(x)| and f(x) have the same roots.
This leaves the function (f(x))–1. We can set it equal to zero to find its roots. Remember that, in general, a–1 = 1/a.
(f(x))–1 = 1/f(x) = 0
Multiply by f(x) on both sides.
1 = 0(f(x)) = 0
Clearly, we have something strange here, because one doesn't equal zero. This tells us that (f(x))–1 can't have the same roots as f(x).
Let's look at an example of why the roots of f(x) and (f(x))–1 won't necessarily be the same. Let f(x) = x. The only root of f(x) would be at the point (0,0). Next, let's look at (f(x))–1 = 1/x. The graph of 1/x won't have any roots, because it has a vertical asymptote at zero (1/x is not defined when x = 0). Thus, f(x) doesn't have the same roots as (f(x))–1. Intuitively, this makes since, because (f(x))–1 essentially takes the reciprocal of the y-values on every point of f(x). But zero doesn't have a reciprocal (because 1/0 isn't defined), so the roots will likely change.
The answer is (f(x))–1 .
Example Question #65 : Algebraic Functions
Let the function f be defined by f(x) = x + z. If f(5) = 9, what is the value of f(3z)?
16
24
12
20
4
16
You use the equation to solve for z, which equals 4. We then plug this in to f(3z), making x = 12, then add 4 as in the original equation, giving you 16.
Example Question #67 : Algebraic Functions
Jorge is studying the graph of the function f(x). He notices that f(x) passes through the points (–1,5), (1,–1), and (2,–16). Which of the following equations could represent f(x)?
f(x) = –1.5x2 – 3x + 3.5
f(x) = 2x2 – 3x3
f(x) = –x3 – 4x
f(x) = –3x + 2
f(x) = 1.5x2 – 3x + 0.5
f(x) = 2x2 – 3x3
We are told that f(x) must pass through the points (–1,5), (1,–1), and (2,–16). In order for it to pass through (–1,5), when x = –1 is input into the equation of f(x), the result must be 5. Likewise, when 1 and 2 are input into the equation of f(x), the results should be –1 and –16, respectively. To check if the equation for f(x) is correct, we need to put x = –1, 1, and 2 into the equation and see if we get the correct results. In short, we need to find the equation of f(x) such that f(–1) = 5, f(1) = –1, and f(2) = –16.
The only equation for f(x) that will pass through all three points is f(x) = 2x2 – 3x3 . We can see this by testing x = –1, 1, and 2.
f(x) = 2x2 – 3x3
f(–1) = 2(–1)2 – 3(–1)3 = 2(1) – 3(–1) = 5
This means that (–1,5) would indeed lie on the graph of f(x).
f(1) = 2(1)2 – 3(1)3 = 2 – 3 = –1
So f(x) would also pass through (1,–1).
f(2) = 2(2)2 – 3(2)3 = 8 – 3(8) = –16
f(x) would also pass through (2,16).
We can check to see why the other equations of f(x) wouldn't satisfy all three points.
Let's look at the equation f(x) = –3x + 2 and test x = 2.
f(2) = –3(2) + 2 = –4, which doesn't equal –16. So that means this equation wouldn't pass through (2,–16).
Next, we can try f(x) = –1.5x2 – 3x + 3.5, using x = 2.
f(2) = –1.5(22) – 3(2) + 3.5 = –8.5, which is not –16.
Next, we can try f(x) = 1.5x2 – 3x + 0.5 at x = 2.
f(2) = 1.5(22) – 3(2) + 0.5 = 0.5, which is not equal to –16.
Finally, let's try f(x) = –x3 – 4x, using x = 1.
f(1) = –(13) – 4(1) = –5, which doesn't equal –1.
The answer is f(x) = 2x2 – 3x3 .
Example Question #26 : How To Find F(X)
Let f(x) = ax2, where a is a positive constant. Which of the following statements must be true for all values of a?
I. f(x) = f(–x)
II. f(x) = |f(x)|
III. f(f(x)) = a3x4
II and III only
I and III only
I, II, and III
I and II only
I only
I, II, and III
Let's go through each statement and determine if it's true.
f(x) = ax2
f(–x) = a(–x)2 = a(–x)(–x) = ax2 = f(x)
Statement I is true. Now let's look at the next statement.
f(x) = ax2
|f(x)| = |ax2|
Because a is positive, and because x2 ≥ 0, the product of a and x2 must be greater than or equal to zero. In other words, ax2 ≥ 0. If we take the absolute value of a quantity that is greater than or equal to zero, we get that quantity. In general, if b ≥ 0, then |b| = b. This means that |ax2| = ax2. Thus, we can write the following:
|f(x)| = |ax2| = ax2 = f(x)
Therefore, statement II is also true. Let's look at the last statement.
f(f(x)) = f(ax2)= a(ax2)2 = a(ax2)(ax2) = a3x4
This means III is also true.
The answer is I, II, and III.
Example Question #2773 : Sat Mathematics
Which relation is NOT a function?
2y + 3x = 6
{(–3, –6), (–2, –1), (–1, 0), (0, 3), (1, 0), (1, 15)}
{(–3, –6), (–2, –6), (–1, –6), (0, –6), (1, –6)}
y = 3x + 1
{(–3, –6), (–2, –1), (–1, 0), (0, 3), (1, 15)}
{(–3, –6), (–2, –1), (–1, 0), (0, 3), (1, 0), (1, 15)}
A relation is a function if every x-value corresponds to one and only one y-value. Let's look at our answer choices.
{(–3, –6), (–2, –1), (–1, 0), (0, 3), (1, 15)}: This is a function. There is only one y for each x-value.
{(–3, –6), (–2, –6), (–1, –6), (0, –6), (1, –6)}: This might not look like a fuction, but it is indeed a function. This set follows the rule that every x maps to only one y-value, it just happens that the y-value (-6) is the same for each x-value.
{(–3, –6), (–2, –1), (–1, 0), (0, 3), (1, 0), (1, 15)}: This is NOT a function and therefore the correct answer. There are two x-values (both 1) that map to two different y-values, (1, 0) and (1, 15). Then this is a relation but NOT a function.
2y + 3x = 6: This is also a function. If you can solve for y, then the equation is a function. Here we can isolate y: y = –3x/2 + 3.
y = 3x + 1: This is similar to the last answer choice, but with even less work to decide this is a function. Clearly this equation is in the form of "y = " and solves for a unique y-value for every x.
Example Question #2774 : Sat Mathematics
For all values of x, there exist two functions, f(x) = 3x2 + 4 and g(x) = 6x – 1. What is g(f(x))?
15x + 4
18x2 + 23
(6x – 1)2
18x2 – 23
15x – 4
18x2 + 23
In the composite function g(f(x)), f(x) acts like "x" or any other variable. We need to plug f(x) into g(x). So g(f(x)) = 6(3x2 + 4) – 1 = 18x2 + 24 – 1 = 18x2 + 23.
Example Question #2775 : Sat Mathematics
Let f(x) be defined as follows:
What is the value of
7
5
15
13
17
17
The function f(x) is a piecewise function, which means that it is comprised of separate functions that change depending on the value of x. According to the problem, whenever x is less than 1, f(x) is defined by 2x - 3. Whenever x is greater than or equal to one and less than 3, f(x) is defined as . And whenever x is greater than or equal to 3, f(x) is equal to |4x – 9|.
The question asks us to find f(1) + f(3) + f(5). We need to calculate the values of f(1), f(3), and f(5) individually and then find their sum.
To find f(1), we must first decide which of the three possible functions for f(x) to use. Since f(1) means we are finding the value of f(x) when x = 1, we will have to use the second piece, which says that f(x) = . We can't use the function 2x - 3, because this is only valid when x < 1, not when x = 1.
Next, we will find f(3). We need to use the third piece of the function which states that f(x) = |4x – 9|. Since f(3) means we are finding f(x) when x = 3, we can only use the third piece. The second piece, , is only defined if x is less than 3 and greater than or equal to 1.
Lastly, we must find f(5). Again, we will use the function f(x) = |4x – 9|, because when x = 5, x must be greater than 3.
We can now add up f(1), f(3), and f(5), which would give us 3 + 3 + 11 = 17.
The answer is 17.
Example Question #72 : Algebraic Functions
If a function, g, is defined by , what is when x = 2?
7
3
25
16
49
25
First, let's expand . We will get the polynomial .
When we plug in the given function, g, into this polynomial and simplify, we will get .
Because this problem asks us for the value when x = 2, we have to plug in 2 for every x in the expression. .