Algebraic Functions - PSAT Math
Card 1 of 826
If the function g is defined by g(x) = 4_x_ + 5, then 2_g_(x) – 3 =
If the function g is defined by g(x) = 4_x_ + 5, then 2_g_(x) – 3 =
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The function g(x) is equal to 4_x_ + 5, and the notation 2_g_(x) asks us to multiply the entire function by 2. 2(4_x_ + 5) = 8_x_ + 10. We then subtract 3, the second part of the new equation, to get 8_x_ + 7.
The function g(x) is equal to 4_x_ + 5, and the notation 2_g_(x) asks us to multiply the entire function by 2. 2(4_x_ + 5) = 8_x_ + 10. We then subtract 3, the second part of the new equation, to get 8_x_ + 7.
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If f(x) = x_2 + 5_x and g(x) = 2, what is f(g(4))?
If f(x) = x_2 + 5_x and g(x) = 2, what is f(g(4))?
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First you must find what g(4) is. The definition of g(x) tells you that the function is always equal to 2, regardless of what “x” is. Plugging 2 into f(x), we get 22 + 5(2) = 14.
First you must find what g(4) is. The definition of g(x) tells you that the function is always equal to 2, regardless of what “x” is. Plugging 2 into f(x), we get 22 + 5(2) = 14.
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f(a) = 1/3(a_3 + 5_a – 15)
Find a = 3.
f(a) = 1/3(a_3 + 5_a – 15)
Find a = 3.
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Substitute 3 for all a.
(1/3) * (33 + 5(3) – 15)
(1/3) * (27 + 15 – 15)
(1/3) * (27) = 9
Substitute 3 for all a.
(1/3) * (33 + 5(3) – 15)
(1/3) * (27 + 15 – 15)
(1/3) * (27) = 9
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Evaluate f(g(6)) given that f(x) = _x_2 – 6 and g(x) = –(1/2)x – 5
Evaluate f(g(6)) given that f(x) = _x_2 – 6 and g(x) = –(1/2)x – 5
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Begin by solving g(6) first.
g(6) = –(1/2)(6) – 5
g(6) = –3 – 5
g(6) = –8
We substitute f(–8)
f(–8) = (–8)2 – 6
f(–8) = 64 – 6
f(–8) = 58
Begin by solving g(6) first.
g(6) = –(1/2)(6) – 5
g(6) = –3 – 5
g(6) = –8
We substitute f(–8)
f(–8) = (–8)2 – 6
f(–8) = 64 – 6
f(–8) = 58
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If f(x) = |(_x_2 – 175)|, what is the value of f(–10) ?
If f(x) = |(_x_2 – 175)|, what is the value of f(–10) ?
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If x = –10, then (_x_2 – 175) = 100 – 175 = –75. But the sign |x| means the absolute value of x. Absolute values are always positive.
|–75| = 75
If x = –10, then (_x_2 – 175) = 100 – 175 = –75. But the sign |x| means the absolute value of x. Absolute values are always positive.
|–75| = 75
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If f(x)= 2x² + 5x – 3, then what is f(–2)?
If f(x)= 2x² + 5x – 3, then what is f(–2)?
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By plugging in –2 for x and evaluating, the answer becomes 8 – 10 – 3 = -5.
By plugging in –2 for x and evaluating, the answer becomes 8 – 10 – 3 = -5.
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If f(x) = x² – 2 and g(x) = 3x + 5, what is f(g(x))?
If f(x) = x² – 2 and g(x) = 3x + 5, what is f(g(x))?
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To find f(g(x) plug the equation for g(x) into equation f(x) in place of “x” so that you have: f(g(x)) = (3x + 5)² – 2.
Simplify: (3x + 5)(3x + 5) – 2
Use FOIL: 9x² + 30x + 25 – 2 = 9x² + 30x + 23
To find f(g(x) plug the equation for g(x) into equation f(x) in place of “x” so that you have: f(g(x)) = (3x + 5)² – 2.
Simplify: (3x + 5)(3x + 5) – 2
Use FOIL: 9x² + 30x + 25 – 2 = 9x² + 30x + 23
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If f(x)=3x and g(x)=2x+2, what is the value of f(g(x)) when x=3?
If f(x)=3x and g(x)=2x+2, what is the value of f(g(x)) when x=3?
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With composition of functions (as with the order of operations) we perform what is inside of the parentheses first. So, g(3)=2(3)+2=8 and then f(8)=24.
With composition of functions (as with the order of operations) we perform what is inside of the parentheses first. So, g(3)=2(3)+2=8 and then f(8)=24.
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The cost of a cell phone plan is $40 for the first 100 minutes of calls, and then 5 cents for each minute after. If the variable x is equal to the number of minutes used for calls in a month on that cell phone plan, what is the equation f(x) for the cost, in dollars, of the cell phone plan for calls during that month?
The cost of a cell phone plan is $40 for the first 100 minutes of calls, and then 5 cents for each minute after. If the variable x is equal to the number of minutes used for calls in a month on that cell phone plan, what is the equation f(x) for the cost, in dollars, of the cell phone plan for calls during that month?
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40 dollars is the constant cost of the cell phone plan, regardless of minute usage for calls. We then add 5 cents, or 0.05 dollars, for every minute of calls over 100. Thus, we do not multiply 0.05 by x, but rather by (x - 100), since the 5 cent charge only applies to minutes used that are over the 100-minute barrier. For example, if you used 101 minutes for calls during the month, you would only pay the 5 cents for that 101st minute, making your cost for calls $40.05. Thus, the answer is 40 + 0.05(x - 100).
40 dollars is the constant cost of the cell phone plan, regardless of minute usage for calls. We then add 5 cents, or 0.05 dollars, for every minute of calls over 100. Thus, we do not multiply 0.05 by x, but rather by (x - 100), since the 5 cent charge only applies to minutes used that are over the 100-minute barrier. For example, if you used 101 minutes for calls during the month, you would only pay the 5 cents for that 101st minute, making your cost for calls $40.05. Thus, the answer is 40 + 0.05(x - 100).
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Which of the following values of x is not in the domain of the function y = (2_x –_ 1) / (x_2 – 6_x + 9) ?
Which of the following values of x is not in the domain of the function y = (2_x –_ 1) / (x_2 – 6_x + 9) ?
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Values of x that make the denominator equal zero are not included in the domain. The denominator can be simplified to (x – 3)2, so the value that makes it zero is 3.
Values of x that make the denominator equal zero are not included in the domain. The denominator can be simplified to (x – 3)2, so the value that makes it zero is 3.
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Given the relation below:
{(1, 2), (3, 4), (5, 6), (7, 8)}
Find the range of the inverse of the relation.
Given the relation below:
{(1, 2), (3, 4), (5, 6), (7, 8)}
Find the range of the inverse of the relation.
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The domain of a relation is the same as the range of the inverse of the relation. In other words, the x-values of the relation are the y-values of the inverse.
The domain of a relation is the same as the range of the inverse of the relation. In other words, the x-values of the relation are the y-values of the inverse.
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f(x) = 4x + 2
g(x) = 3x - 1
The two equations above define the functions f(x) = g(x). If f(d) = 2g(d) for some value of d, then what is the value of d?
f(x) = 4x + 2
g(x) = 3x - 1
The two equations above define the functions f(x) = g(x). If f(d) = 2g(d) for some value of d, then what is the value of d?
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f(x) = 4x + 2
g(x) = 3x - 1
We have f(d) = 2g(d). We multiply each value in g(d) by 2.
4d + 2 = 2(3d - 1) (Distribute the 2 in the parentheses by multiplying each value in them by 2.)
4d + 2 = 6d - 2 (Add 2 to both sides.)
4d + 4 = 6d (Subtract 4d from both sides.)
4 = 2d (Divide both sides by 2.)
2 = d
We can plug that back in to double check.
4(2) + 2 = 6(2) - 2
8 + 2 = 12 - 2
10 = 10
f(x) = 4x + 2
g(x) = 3x - 1
We have f(d) = 2g(d). We multiply each value in g(d) by 2.
4d + 2 = 2(3d - 1) (Distribute the 2 in the parentheses by multiplying each value in them by 2.)
4d + 2 = 6d - 2 (Add 2 to both sides.)
4d + 4 = 6d (Subtract 4d from both sides.)
4 = 2d (Divide both sides by 2.)
2 = d
We can plug that back in to double check.
4(2) + 2 = 6(2) - 2
8 + 2 = 12 - 2
10 = 10
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If 
, then which of the following is equal to 
?
If
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What is the range of the function y = _x_2 + 2?
What is the range of the function y = _x_2 + 2?
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The range of a function is the set of y-values that a function can take. First let's find the domain. The domain is the set of x-values that the function can take. Here the domain is all real numbers because no x-value will make this function undefined. (Dividing by 0 is an example of an operation that would make the function undefined.)
So if any value of x can be plugged into y = _x_2 + 2, can y take any value also? Not quite! The smallest value that y can ever be is 2. No matter what value of x is plugged in, y = _x_2 + 2 will never produce a number less than 2. Therefore the range is y ≥ 2.
The range of a function is the set of y-values that a function can take. First let's find the domain. The domain is the set of x-values that the function can take. Here the domain is all real numbers because no x-value will make this function undefined. (Dividing by 0 is an example of an operation that would make the function undefined.)
So if any value of x can be plugged into y = _x_2 + 2, can y take any value also? Not quite! The smallest value that y can ever be is 2. No matter what value of x is plugged in, y = _x_2 + 2 will never produce a number less than 2. Therefore the range is y ≥ 2.
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What is the smallest value that belongs to the range of the function 
?
What is the smallest value that belongs to the range of the function
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We need to be careful here not to confuse the domain and range of a function. The problem specifically concerns the range of the function, which is the set of possible numbers of 
. It can be helpful to think of the range as all the possible y-values we could have on the points on the graph of 
.
Notice that 
has 
in its equation. Whenever we have an absolute value of some quantity, the result will always be equal to or greater than zero. In other words, |4-x| 
0. We are asked to find the smallest value in the range of 
, so let's consider the smallest value of 
, which would have to be zero. Let's see what would happen to 
if 
.
This means that when 
, 
. Let's see what happens when 
gets larger. For example, let's let 
.
As we can see, as 
gets larger, so does 
. We want 
to be as small as possible, so we are going to want 
to be equal to zero. And, as we already determiend, 
equals 
when 
.
The answer is 
.
We need to be careful here not to confuse the domain and range of a function. The problem specifically concerns the range of the function, which is the set of possible numbers of
Notice that
This means that when
As we can see, as
The answer is
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If 
, then find 
If
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is the same as 
.
To find the inverse simply exchange 
and 
and solve for 
.
So we get 
which leads to 
.
To find the inverse simply exchange
So we get
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What is the range of the function y = _x_2 + 2?
What is the range of the function y = _x_2 + 2?
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The range of a function is the set of y-values that a function can take. First let's find the domain. The domain is the set of x-values that the function can take. Here the domain is all real numbers because no x-value will make this function undefined. (Dividing by 0 is an example of an operation that would make the function undefined.)
So if any value of x can be plugged into y = _x_2 + 2, can y take any value also? Not quite! The smallest value that y can ever be is 2. No matter what value of x is plugged in, y = _x_2 + 2 will never produce a number less than 2. Therefore the range is y ≥ 2.
The range of a function is the set of y-values that a function can take. First let's find the domain. The domain is the set of x-values that the function can take. Here the domain is all real numbers because no x-value will make this function undefined. (Dividing by 0 is an example of an operation that would make the function undefined.)
So if any value of x can be plugged into y = _x_2 + 2, can y take any value also? Not quite! The smallest value that y can ever be is 2. No matter what value of x is plugged in, y = _x_2 + 2 will never produce a number less than 2. Therefore the range is y ≥ 2.
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If f(x) = _x_2 – 5 for all values x and f(a) = 4, what is one possible value of a?
If f(x) = _x_2 – 5 for all values x and f(a) = 4, what is one possible value of a?
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Using the defined function, f(a) will produce the same result when substituted for x:
f(a) = _a_2 – 5
Setting this equal to 4, you can solve for a:
_a_2 – 5 = 4
_a_2 = 9
a = –3 or 3
Using the defined function, f(a) will produce the same result when substituted for x:
f(a) = _a_2 – 5
Setting this equal to 4, you can solve for a:
_a_2 – 5 = 4
_a_2 = 9
a = –3 or 3
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The function 
is defined as 
. What is 
?
The function 
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Substitute -1 for 
in the given function.
If you didn’t remember the negative sign, you will have calculated 36. If you remembered the negative sign at the very last step, you will have calculated -36; however, if you did not remember that 
is 1, then you will have calculated 18.
Substitute -1 for 
If you didn’t remember the negative sign, you will have calculated 36. If you remembered the negative sign at the very last step, you will have calculated -36; however, if you did not remember that 
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f(x) = 2x2 + x – 3 and g(y) = 2y – 7. What is f(g(4))?
f(x) = 2x2 + x – 3 and g(y) = 2y – 7. What is f(g(4))?
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To evaluate f(g(4)), one must first determine the value of g(4), then plug that into f(x).
g(4) = 2 x 4 – 7 = 1.
f(1) = 2 x 12 + 2 x 1 – 3 = 0.
To evaluate f(g(4)), one must first determine the value of g(4), then plug that into f(x).
g(4) = 2 x 4 – 7 = 1.
f(1) = 2 x 12 + 2 x 1 – 3 = 0.
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