Which of the following functions are one-to-one? Select all that apply.

Question

a) lower f left parenthesis x right parenthesis equals x superscript 3 baseline minus 7Image with alt text: lower f left parenthesis x right parenthesis equals x superscript 3 baseline minus 7

b) lower f left parenthesis x right parenthesis equals x superscript 2 baseline minus 4
Image with alt text: lower f left parenthesis x right parenthesis equals x superscript 2 baseline minus 4

c) lower f left parenthesis x right parenthesis equals Start Fraction 1 over 8 x minus 1 End Fraction
Image with alt text: lower f left parenthesis x right parenthesis equals Start Fraction 1 over 8 x minus 1 End Fraction

d) lower f left parenthesis x right parenthesis equals Start Fraction 5 over x superscript 4 baseline End Fraction
Image with alt text: lower f left parenthesis x right parenthesis equals Start Fraction 5 over x superscript 4 baseline End Fraction

e) lower f left parenthesis x right parenthesis equals Start Absolute Value x End Absolute Value

Answer 1

To determine if a function is one-to-one, we need to check if every value in the domain corresponds to a unique value in the range. In other words, if two different inputs produce the same output, the function is not one-to-one.

a) The function f(x) = x^3 - 7 is one-to-one because no two different inputs will produce the same output.

b) The function f(x) = x^2 - 4 is not one-to-one because if we take x = 2 and x = -2, both will produce the same output f(2) = f(-2) = 0.

c) The function f(x) = (1/8)x - 1 is one-to-one because no two different inputs will produce the same output.

d) The function f(x) = 5/x^4 is one-to-one because no two different inputs will produce the same output.

e) The function f(x) = |x| is not one-to-one because both x and -x will produce the same output f(x) = f(-x).

Therefore, the functions that are one-to-one are a), c), and d).

Answer 2

only 2 answers are allowed

Answer 3

In that case, the two functions that are one-to-one are a) and c).

Answer 4

Find the inverse function for lower f left parenthesis x right parenthesis equals left parenthesis x plus 1 right parenthesis superscript 3 baseline minus 2.

Answer 5

To find the inverse function for f(x) = (x + 1)^3 - 2, we need to swap the roles of x and f(x) and solve for the new f(x), which will be the inverse function.

Let y = (x + 1)^3 - 2

Swap x and y:

x = (y + 1)^3 - 2

Now, solve for y:

x = (y + 1)^3 - 2
x + 2 = (y + 1)^3
∛(x + 2) = y + 1
∛(x + 2) - 1 = y

Therefore, the inverse function of f(x) = (x + 1)^3 - 2 is:

f^(-1)(x) = ∛(x + 2) - 1

Answer 6

Determine the domain and range for the inverse of lower f left parenthesis x right parenthesis equals Start Fraction 1 over x End Fraction plus 5.

Answer 7

To determine the domain and range of the inverse function of f(x) = 1/x + 5, we need to consider the restrictions and possible values for the input and output.

First, let's find the inverse function of f(x):

y = 1/x + 5

Swap x and y:

x = 1/y + 5

Solve for y:

x - 5 = 1/y

1/y = x - 5

Invert both sides:

y = 1/(x - 5)

The inverse function of f(x) is:

f^(-1)(x) = 1/(x - 5)

Now, let's determine the domain and range of f^(-1)(x):

Domain of f^(-1)(x) is the set of all possible values for x that result in a meaningful output. In this case, the denominator (x - 5) cannot be zero, since division by zero is undefined. Therefore, the domain is:

Domain: x ≠ 5

Range of f^(-1)(x) is the set of all possible values for y (or x in this case) that result from the function. Since the function is always positive (due to 1 divided by a positive value), the range is:

Range: y > 0

Therefore, the domain of the inverse function is x ≠ 5 and the range is y > 0.

Answer 8

choices
domain: left parenthesis negative infinity comma negative 5 right parenthesis union left parenthesis negative 5 comma infinity right parenthesis
range: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis

domain: Image with alt text: left parenthesis negative infinity comma negative 5 right parenthesis union left parenthesis negative 5 comma infinity right parenthesis range: Image with alt text: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis

domain: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis
range: left parenthesis negative infinity comma negative 5 right parenthesis union left parenthesis negative 5 comma infinity right parenthesis

domain: Image with alt text: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis range: Image with alt text: left parenthesis negative infinity comma negative 5 right parenthesis union left parenthesis negative 5 comma infinity right parenthesis

domain: left parenthesis negative infinity comma 5 right parenthesis union left parenthesis 5 comma infinity right parenthesis
range: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis

domain: Image with alt text: left parenthesis negative infinity comma 5 right parenthesis union left parenthesis 5 comma infinity right parenthesis range: Image with alt text: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis

domain: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis
range: left parenthesis negative infinity comma 5 right parenthesis union left parenthesis 5 comma infinity right parenthesis

Answer 9

The correct answer is:

Domain: (-∞, -5) ∪ (-5, ∞)
Range: (-∞, 0) ∪ (0, ∞)

The correct description for the domain is: "Domain: (-∞, -5) ∪ (-5, ∞)" and for the range is "Range: (-∞, 0) ∪ (0, ∞)".