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The function f is defined by

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by jack0997 » Tue Jun 20, 2017 10:30 pm
The function f is defined by f „(x) =… √x - 20 for all positive numbers x. If p ƒ= f (q) for some positive numbers q and p, what is q in terms of p?

(A) (p +‚ 20)^2
(B) √(p + 20‚)
(C) (√p + 20)^2
(D) √(p^2 +‚ 20)
(E) (p^2 +‚ 20)^2

OA A

Pl. help me out with this question,
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by Jay@ManhattanReview » Tue Jun 20, 2017 10:54 pm
jack0997 wrote:The function f is defined by f „(x) =… √x - 20 for all positive numbers x. If p ƒ= f (q) for some positive numbers q and p, what is q in terms of p?

(A) (p +‚ 20)^2
(B) √(p + 20‚)
(C) (√p + 20)^2
(D) √(p^2 +‚ 20)
(E) (p^2 +‚ 20)^2

OA A

Pl. help me out with this question,
„f(x)… ƒ= √x - 20

=> f(q) = √q - 20

p = √q - 20

=> √q = p + 20

Squaring both sides, we have:

=> q =[spoiler] (p + 20)^2[/spoiler]

The correct answer: A

Hope this helps!

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by [email protected] » Wed Jun 21, 2017 2:13 pm
Hi jack0997,

This question can be solved by TESTing VALUES.

Based on the information in the prompt, my initial thought would be to TEST Q=400....

f(400) = 20 - 20 = 0 and that would give us P=0.

This is a problem for a couple of reasons: First, 0 is NOT positive - and we're told that P and Q have to be positive. Second, you'll notice that three of the answers (A, C and E) would all end in a 400 when you plug in P=0 (and there can't be three correct answers here). Thus, we have to look for a different 'round number' for Q, but one that's bigger than 20.

Let's TEST Q = 900

f(900) = 30 - 20 = 10, so we're looking for an answer that equals 900 when we plug in P=10. You'll find that only one answer matches...

Final Answer: A

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by Matt@VeritasPrep » Thu Jun 22, 2017 4:52 pm
p = √q - 20

p + 20 = √q

(p + 20)² = q
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