Algebraic Functions - PSAT Math

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.

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.

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 .

is the same as .

To find the inverse simply exchange and and solve for .

So we get which leads to .

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.

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

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.

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.

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.

There are two ways to do this problem. The first way just involves plugging in –3 for x and solving 6〖(–3)〗2 + 16(–3) – 6, which equals 54 – 48 – 6 = 0. The second way involves factoring the polynomial to (6x – 2)(x + 3) and then plugging in –3 for x. The second way quickly shows that the answer is 0 due to multiplying by (–3 + 3).