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The function f is defined by

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by jack0997 » Tue Jun 20, 2017 10:16 pm
The function f is defined by f „(x) =… ƒ –1/x for all non-zero numbers x. If f „(a) …= ƒ–1/2 and f „(ab)… = ƒ1/6, then b ƒ=

(A) 3
(B)1/3
(C) –1/3
(D) –3
(E) –12

OA D

Pl. help me out with this question.
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by Jay@ManhattanReview » Tue Jun 20, 2017 10:48 pm
jack0997 wrote:The function f is defined by f „(x) =… ƒ –1/x for all non-zero numbers x. If f „(a) …= ƒ–1/2 and f „(ab)… = ƒ1/6, then b ƒ=

(A) 3
(B)1/3
(C) –1/3
(D) –3
(E) –12

OA D

Pl. help me out with this question.
f „(x)… ƒ = –1/x

=ƒ> f „(a)… ƒ = –1/a = –1/2
ƒ=> a ƒ= 2

Also, we have:
f „(ab)… ƒ = –1/ab = 1/6
=> ab = –6
=> b = –6/a = –6/2

=> [spoiler]b = –3[/spoiler]

The correct answer: D

Hope this helps!

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by Matt@VeritasPrep » Thu Jun 22, 2017 5:04 pm
The function f is defined by f „(x) =… ƒ -1/x for all non-zero numbers x. If f „(a) …= ƒ-1/2 and f „(ab)… = ƒ1/6, then b ƒ=

f(a) = -1/2
and, by definition,
f(a) = -1/a

so a = 2

f(ab) = f(2b) = 1/6
and, by definition
f(2b) = -1/2b

so b = -3
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