Question

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Find the inverse of the given function.
27pionts

Answer 1

Question One: log_2(x - 6) = y

Question Two: First blank: 4 Second blank 0

Question Three: B

Explanation:

f(x) = 2^x + 6

y = 2^x + 6 Interchange the x and y

x = 2^y + 6 Subtract 6 from both sides

x - 6 = 2^y Take the log of both sides

log(x - 6) = log 2^y The power can be multiplied by the log of 2

log(x - 6) = y log(2) Divide by the log of 2

log(x - 6)/log(2) = y This to me is the preferred answer. It allows you to calculate what the actual number is or graph it on Desmos

The answer I have given is equivalent to using base 2 as your log, but calculating that is not always easy.

log_2(x - 6) = y

Answer: D

Problem Two

You want the inverse of y = - 1/2 sqrt(x + 3) x ≥ - 3

Note: you cannot have any number less than -3 because if you do, you will be taking the square root of -x which involves complex numbers. In addition x ≥ -3 is a domain. You have to keep that in mind when doing this question.

• y = - 1/2 sqrt(x + 3) Interchange the x and y
• x = - 1/2 sqrt(y + 3) Multiply by - 2
• -2x = sqrt(y + 3) Square both sides
• (-2x)^2 = sqrt(y + 3)^2
• 4x^2 = y + 3 Subtract 3 from both sides
• y = 4x^2 - 3

The domain and range of this is a little harder to figure out. The range of f(x) becomes the domain of f-1(x)

The range of f(x) is 0 <=y < - infinity

So the domain of f-1(x) <= 0

Answer: The first blank is 4 and the second one is 0

Graph: the graph is given you below the question. What it shows is the f(x) and f^-1(x) are symmetrical about y = x. Most inverses are. If you want to check your work, it is best to include a graph when doing inverses.

• Red Original
• Blue inverse
• Black: y = x

Problem Three

The best way to begin this problem is to apply the associative property of multiplication and start by rewriting the givens as x^(1/2)[(x - 6)(x + 3)]. Do what is in the square brackets first.

• x^(1/2) [x^2 - 6x + 3x - 18]
• x^(1/2) [x^2 - 3x - 18]

Now deal with the x^(1/2) which must be multiplied by all three terms in the square brackets.

• x^(1/2)*x^2: x^(2 + 1/2) = x^(4/2 + 1/2) = x^(5/2)
• x^(1/2)*(-3x): - 3 x^(2/2 + 1/2) = - 3x^(3/2)
• x^(1/2)*(-18): - 18x^(1/2)

Result: x^(5/2) - 3x^(3/2) - 18x^(1/2)

Answer: √(x^5) - 3√x^3 - 18√x

Answer: B

Answer 2

1. D. f^-1(x) = log2(x -6)
2. 4x^2 -3 . . . . x ≤ 0
3. B. √(x^5) -3√(x^3) -18√x

Explanation:

1. When you replace f(x) by x and x by y, you have

... x = 2^y + 6

The first thing you do is subtract 6; then you take the base-2 logarithm:

... (x -6) = 2^y

... log2(x -6) = y = f^-1(x)

You know that to get the y-term by itself, you must subtract 6. Anything else you do will operate on (x-6). Only answer choice D has that sort of construction.

2. When you swap x and y and solve for y, you have ...

... x = -1/2√(y+3)

... -2x = √(y +3) . . . . . . multiply by -2

... (-2x)^2 = y +3 . . . . . square

... 4x^2 - 3 = y = f^-1(x) . . . . subtract 3

The range of f(x) is (-∞, 0], so that is the domain of f^-1(x). That is, f^-1(x) is defined for x ≤ 0.

3. The product of the two functions is ...

... (x -6)(√x)(x +3) = (√x)(x^2 -3x -18)

Every term will have a factor of √x, and the coefficients will be 1, -3, -18. Only selection B matches those conditions.