How to find domain and range of the inverse of a relation

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PSAT Math › How to find domain and range of the inverse of a relation

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What is the range of the function y = _x_2 + 2?

all real numbers

y ≥ 2

CORRECT

{–2, 2}

{2}

undefined

Explanation

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.

What is the smallest value that belongs to the range of the function $f(x)=2|4-x|-2$ ?

$-2$

CORRECT

$-4$

$0$

$2$

$4$

Explanation

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 $f(x)$ . 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 $f(x)$ .

Notice that $f(x)$ has $|4-x|$ 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| $\geq$ 0. We are asked to find the smallest value in the range of $f(x)$ , so let's consider the smallest value of $|4-x|$ , which would have to be zero. Let's see what would happen to $f(x)$ if $|4-x|=0$ .

$f(x)=2(0)-2=0-2=-2$

This means that when $|4-x|=0$ , $f(x)=-2$ . Let's see what happens when $|4-x|$ gets larger. For example, let's let $|4-x|=3$ .

$f(x)=2(3)-2=4$

As we can see, as $|4-x|$ gets larger, so does $f(x)$ . We want $f(x)$ to be as small as possible, so we are going to want $|4-x|$ to be equal to zero. And, as we already determiend, $f(x)$ equals $-2$ when $|4-x|=0$ .

The answer is $-2$ .

Given the relation below, identify the domain of the inverse of the relation.

$\left \{(1, 2), (3, 4), (5, 6), (7, 8){ \right \}$

$\left \{ 1,\ 2,\ 3,\ 4 \right \}$

$\left \{ 5,\ 6,\ 7,\ 8 \right \}$

$\left \{ 1,\ 3,\ 5,\ 7 \right \}$

$\left \{ 2,\ 4,\ 6,\ 8 \right \}$

CORRECT

The inverse of the relation does not exist.

Explanation

$\left \{(1, 2), (3, 4), (5, 6), (7, 8){ \right \}$

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: $\left \{ 2,\ 4,\ 6,\ 8 \right \}$ .

Thus, the domain for the inverse relation will also be $\left \{ 2,\ 4,\ 6,\ 8 \right \}$ .

If $f(x)=\frac{x}{x-3}$ , then which of the following is equal to $f^{-1}(x)$ ?

$\frac{-x}{x+3}$

$\frac{-x}{x-3}$

$\frac{3x}{x-1}$

CORRECT

$\frac{-3x}{x+1}$

Explanation

Inversef2

Inverse3

What is the range of the function y = _x_2 + 2?

y ≥ 2

CORRECT

all real numbers

{–2, 2}

{2}

undefined

Explanation

Which of the following values of x is not in the domain of the function y = (2_x –_ 1) / (x_2 – 6_x + 9) ?

–1/2

1/2

CORRECT

Explanation

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.

Given the relation below:

{(1, 2), (3, 4), (5, 6), (7, 8)}

Find the range of the inverse of the relation.

{1, 2, 3, 4}

{5, 6, 7, 8}

{1, 3, 5, 7}

CORRECT

{2, 4, 6, 8}

the inverse of the relation does not exist

Explanation

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.

If $f(x) = x - 3$ , then find $f^{-1}(x)$