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SAT Math : How to find domain and range of the inverse of a relation

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Example Questions

Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

Which of the following values of x is not in the domain of the function y = (2x – 1) / (x² – 6x + 9) ?

Possible Answers:

–1/2

1/2

Correct answer:

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.

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Example Question #2 : How To Find Domain And Range Of The Inverse Of A Relation

Given the relation below:

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

Find the range of the inverse of the relation.

Possible Answers:

{5, 6, 7, 8}

{1, 2, 3, 4}

{1, 3, 5, 7}

{2, 4, 6, 8}

the inverse of the relation does not exist

Correct answer: {1, 3, 5, 7}

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.

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Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

What is the range of the function y = x² + 2?

Possible Answers:

y ≥ 2

{–2, 2}

all real numbers

{2}

undefined

Correct answer:

y ≥ 2

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 = x² + 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 = x² + 2 will never produce a number less than 2. Therefore the range is y ≥ 2.

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Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

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

Possible Answers:

$\dpi{100} -2$

$\dpi{100} 0$

$\dpi{100} 2$

$\dpi{100} 4$

$\dpi{100} -4$

Correct answer:

$\dpi{100} -2$

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 $\dpi{100} f(x)$ . 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 $\dpi{100} f(x)$ .

Notice that $\dpi{100} f(x)$ has $\dpi{100} |4-x|$ 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 $\dpi{100} f(x)$ , so let's consider the smallest value of $\dpi{100} |4-x|$ , which would have to be zero. Let's see what would happen to $\dpi{100} f(x)$ if $\dpi{100} |4-x|=0$ .

$\dpi{100} f(x)=2(0)-2=0-2=-2$

This means that when $\dpi{100} |4-x|=0$ , $\dpi{100} f(x)=-2$ . Let's see what happens when $\dpi{100} |4-x|$ gets larger. For example, let's let $\dpi{100} |4-x|=3$ .

$\dpi{100} f(x)=2(3)-2=4$

As we can see, as $\dpi{100} |4-x|$ gets larger, so does $\dpi{100} f(x)$ . We want $\dpi{100} f(x)$ to be as small as possible, so we are going to want $\dpi{100} |4-x|$ to be equal to zero. And, as we already determiend, $\dpi{100} f(x)$ equals $\dpi{100} -2$ when $\dpi{100} |4-x|=0$ .

The answer is $\dpi{100} -2$ .

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Example Question #3 : How To Find Domain And Range Of The Inverse Of A Relation

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

Possible Answers:

$f^{-1}(x)=3-x$

$f^{-1}(x)=x+3$

$f^{-1}(x)=3x$

$f^{-1}(x)=x-3$

$f^{-1}(x)=\frac{1}{x-3}$

Correct answer:

$f^{-1}(x)=x+3$

Explanation:

$f(x) = x - 3$ is the same as $y= x - 3$ .

To find the inverse simply exchange $x$ and $y$ and solve for $y$ .

So we get $x=y-3$ which leads to $y=x+3$ .

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Example Question #2 : How To Find Domain And Range Of The Inverse Of A Relation

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Inversef2

Inverse3

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Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

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

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

Possible Answers:

The inverse of the relation does not exist.

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

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

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

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

Correct answer:

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

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

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Example Question #5 : How To Find Domain And Range Of The Inverse Of A Relation

Define $f(x) = x^{2}- 4x - 4$ , restricting the domain of the function to $\left ( -\infty , 0 \right ]$ .

Determine $f^{-1} (x)$ (you need not determine its domain restriction).

Possible Answers:

$f^{-1}(x)= 2 - \sqrt{ x+ 8 }$

$f^{-1} (x)$ does not exist

$f^{-1}(x)= 2 - \sqrt{ x-8 }$

$f^{-1}(x)= 2 + \sqrt{ x-8 }$

$f^{-1}(x)= 2 + \sqrt{ x+ 8 }$

Correct answer:

$f^{-1}(x)= 2 - \sqrt{ x+ 8 }$

Explanation:

First, we must determine whether $f^{-1} (x)$ exists.

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place.

$f^{-1} (x)$ exists if and only if, if f(a) = f(b) , then a = b - or, equivalently, if there does not exist and such that $a \ne b$ , but . This will happen on any interval on which the graph of constantly increases or constantly decreases, but if the graph changes direction on an interval, there will be $a \ne b$ such that f(a) = f(b) on this interval. The key is therefore to determine whether the interval to which the domain is restricted contains the vertex.

The -coordinate of the vertex of the parabola of the function

$f(x) = ax^{2}+ bx+c$

is $x = - \frac{b}{2a}$ .

The -coordinate of the vertex of the parabola of $f(x) = x^{2}- 4x - 4$ can be found by setting a = 1, b = -4 :

$x = - \frac{-4}{2 \cdot 1} = \frac{4}{2} =2$ .

The vertex of the graph of without its domain restriction is at the point with -coordinate 2. However, $2 \notin \left ( -\infty , 0 \right ]$ . Therefore, the domain at which f(x) is restricted does not include the vertex, and $f^{-1} (x)$ exists on this domain.

To determine the inverse of f(x) , first, rewrite f(x) in vertex form

f(x) = a (x-h)+k

a = 1 , the same as in the standard form.

The graph of , if unrestricted, would have a vertex with -coordinate 2, and -coordinate

$f(2) = 2^{2}- 4 \cdot 2 - 4 = 4 - 8 - 4 =-8$ .

Therefore, h = 2, k=8 .

The vertex form of f(x) is therefore

$f(x)= (x - 2)^{2}- 8$

To find $f^{-1}(x)$ , first replace f(x) with :

$y= (x - 2)^{2}- 8$

Switch and :

$x= (y - 2)^{2}- 8$

Solve for . First, add 8 to both sides:

$x+ 8 = (y - 2)^{2}- 8 + 8$

$x+ 8 = (y - 2)^{2}$

Take the square root of both sides:

$y - 2 = \pm \sqrt{ x+ 8 }$

Add 2 to both sides

$y - 2+ 2 = \pm \sqrt{ x+ 8 } + 2$

$y = 2 \pm \sqrt{ x+ 8 }$

Replace with $f^{-1}(x)$ :

$f^{-1}(x)= 2 \pm \sqrt{ x+ 8 }$

Either $f^{-1}(x)= 2 + \sqrt{ x+ 8 }$ or $f^{-1}(x)= 2 - \sqrt{ x+ 8 }$

The domain of f(x) is the set of nonpositive numbers; this is consequently the range of $f^{-1}(x)$ . $2 + \sqrt{ x+ 8 }$ can only have positive values, so the only possible choice for $f^{-1}(x)$ is $f^{-1}(x)= 2 - \sqrt{ x+ 8 }$ .

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Example Question #5 : How To Find Domain And Range Of The Inverse Of A Relation

Define a function $f(x) = x^{2} + 9 x -17$ .

It is desired that is domain be restricted so that have an inverse. Which of these domain restrictions would not achieve that goal?

Possible Answers:

$x \in [3,9]$

$x \in [-15, -9]$

$x \in [9, 15 ]$

$x \in [-3, 3]$

$x \in [-9,- 3]$

Correct answer:

$x \in [-9,- 3]$

Explanation:

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place.

The -coordinate of the vertex of the parabola of the function

$f(x) = ax^{2}+ bx+c$

is $x = - \frac{b}{2a}$ .

The -coordinate of the vertex of the parabola of $f(x) = x^{2} + 9 x -17$ can be found by setting a = 1, b = 9 :

$x = - \frac{9}{2 \cdot 1} = - \frac{9}{2} = -4 \frac{1}{2}$ .

Of the five intervals in the choices,

$-4 \frac{1}{2} \in [-9,- 3]$ ,

so $f ^{-1}$ cannot exist if is restricted to this interval. This is the correct choice.

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Example Question #41 : Algebraic Functions

Define $f(x) = \frac{1}{2}x^{2}- 8x - 17$ , restricting the domain of the function to $\left [ 0, \infty \right )$ .

Determine $f^{-1} (x)$ (you need not determine its domain restriction).

Possible Answers:

$f^{-1} (x)$ does not exist

$f^{-1} (x) = 8- \sqrt{2x+98}$

$f^{-1} (x) = 8- \sqrt{2x-98}$

$f^{-1} (x) = 8+ \sqrt{2x-98}$

$f^{-1} (x) = 8+ \sqrt{2x+98}$

Correct answer:

$f^{-1} (x)$ does not exist

Explanation:

First, we must determine whether $f^{-1} (x)$ exists.

A quadratic function has a parabola as its graph; this graph changes direction (downward to upward, or vice versa) at a given point called the vertex.

$f^{-1} (x)$ exists on a given domain interval if and only if there does not exist and on this domain such that $a \ne b$ , but f(a) = f(b) . This will happen if the graph changes direction on the domain interval. The key is therefore to determine whether the given domain interval includes the vertex.

The -coordinate of the vertex of the parabola of the function

$f(x) = ax^{2}+ bx+c$

is $x = - \frac{b}{2a}$ .

The -coordinate of the vertex of the parabola of $f(x) = \frac{1}{2}x^{2}- 8x - 17$ can be found by setting $a = \frac{1}{2},b= -8$ :

$x = - \frac{(-8)}{2\left ( \frac{1}{2} \right )} = -\frac{-8}{1} = 8$ .

The vertex of the graph of without its domain restriction is at the point with -coordinate 8. Since $8 \in \left [ 0, \infty \right )$ , the vertex is in the interior of the domain; as a consequence, the graph of f(x) changes direction on the interval, and $f^{-1} (x)$ does not exist on $\left [ 0, \infty \right )$ .

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SAT Math : How to find domain and range of the inverse of a relation

Study concepts, example questions & explanations for SAT Math

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Example Questions

Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #2 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #3 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #2 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #1 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #5 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #5 : How To Find Domain And Range Of The Inverse Of A Relation

Example Question #41 : Algebraic Functions

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