Algebraic Functions - SAT Math

f(x, y) is defined as x2y2 – xy + y. In order to find f(a, b), we will need to first find a and then b.

We are told that a = f(1, 3). We can use the definition of f(x, y) to determine the value of a.

a = f(1, 3) = 1232 – 1(3) + 3 = 1(9) – 3 + 3 = 9 + 0 = 9

a = 9

Similarly, we can find b by determining the value of f(–2, –1).

b = f(–2, –1) = (–2)2(–1)2 – (–2)(–1) + –1 = 4(1) – (2) – 1 = 4 – 2 – 1 = 1

b = 1

Now, we can find f(a, b), which is equal to f(9, 1).

f(a, b) = f(9, 1) = 92(12) – 9(1) + 1 = 81 – 9 + 1 = 73

f(a, b) = 73

The answer is 73.

F(G(x)) is a composite function where the expression G(x) is substituted in for x in F(x)

F(G(x)) = (x + 5)3 + 2(x + 5)2 – 3 = x_3 + 17_x_2 + 95_x + 172

G(F(x)) = _x_3 + _x_2 + 2

F(x) – G(x) = _x_3 + 2_x_2 – x – 8

F(x) + G(x) = _x_3 + 2_x_2 + x + 2

Evaluating yields –6174.

–147(–21 + 63) =

–147 * 42 = –6174

Subtract the first expression from the second. That gives you 6z = 6. That simplifies to z = 1.

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 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.

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.

Substitute 3 for all a.

(1/3) * (33 + 5(3) – 15)

(1/3) * (27 + 15 – 15)

(1/3) * (27) = 9

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

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

By plugging in –2 for x and evaluating, the answer becomes 8 – 10 – 3 = -5.

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

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.

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).

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

Doing things in an orderly way is a friend to the test-taker.

g(3) = 5 f(3-2)

= 5 f(1)

= 5 \[ 12 + 6**∙**1 + 8\]

= 5 \[ 1 + 6 + 8\]

= 5 \[ 15\]

= 75

For F(1/2), x2=1/4, which is less than 1, so we use the bottom equation to solve. This gives F(1/2)= 4(1/2)2 – 2(1/2) + 2 = 1 – 1 + 2 = 2

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.

The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.

For the original relation, the range is: .

Thus, the domain for the inverse relation will also be .

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.