Answer:
- D. f^-1(x) = log2(x -6)
- 4x^2 -3 . . . . x ≤ 0
- B. √(x^5) -3√(x^3) -18√x
Step-by-step 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)
<em>Every term</em> will have a factor of √x, and the coefficients will be 1, -3, -18. Only selection B matches those conditions.
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