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If f(x) = (x + 6)^2 and g(x) = 9x, which of the following

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If f(x) = (x + 6)^2 and g(x) = 9x, which of the following [#permalink] New post   29 Oct 2020, 11:15
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If f(x) = (x + 6)^2 and g(x) = 9x, which of the following specifies all the possible values of x for which f(g(x)) < g(f(x)) ?

(A) –2 < x < 2

(B) x > 2

(C) 0 < x < 3

(D) –4 < x < 2

(E) 6 < x < 9
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A
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Re: If f(x) = (x + 6)^2 and g(x) = 9x, which of the following [#permalink] New post   29 Oct 2020, 13:32
\(f(g(x)= (9x+6)^2 = 9(3x+2)^2\)

\(g(f(x) = 9(x+6)^2\)

\(f(g(x)) < g(f(x))\)

\(9(3x+2)^2 < 9(x+6)^2\)

\(9x^2+12x+4< x^2+12x+36\)

\(8x^2< 32\)

\(x^2-2^2<0\)

\(-2<x<2\)

Sajjad1994 wrote:
If f(x) = (x + 6)^2 and g(x) = 9x, which of the following specifies all the possible values of x for which f(g(x)) < g(f(x)) ?

(A) –2 < x < 2

(B) x > 2

(C) 0 < x < 3

(D) –4 < x < 2

(E) 6 < x < 9
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Re: If f(x) = (x + 6)^2 and g(x) = 9x, which of the following [#permalink] New post   29 Oct 2020, 13:40
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Sajjad1994 wrote:
If f(x) = (x + 6)^2 and g(x) = 9x, which of the following specifies all the possible values of x for which f(g(x)) < g(f(x)) ?

(A) –2 < x < 2

(B) x > 2

(C) 0 < x < 3

(D) –4 < x < 2

(E) 6 < x < 9




f(x) = \((x + 6)^2\) and g(x) = 9x

f(g(x) means f of g(x) i.e 9x goes into f(x) and becomes \((9x + 6)^2\)

g(f(x) means g of f(x) i.e \((x + 6)^2\) goes into g(x) and becomes \(9(x + 6)^2\)

We want f(g(x) < g(f(x)

\((9x + 6)^2\) < \(9(x + 6)^2\)

\(81x^2 + 108x + 36 < 9(x^2 + 12x + 36)\)

\(81x^2 + 108x + 36 < 9x^2 + 108x + 324)\)

\(81x^2 - 9x^2 < 324 - 36\)

\(72x^2 < 288\)

\(x^2 < 4\)

This means that x can be < 2 or greater than -2 i.e -2 < x < 2


Option A

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Re: If f(x) = (x + 6)^2 and g(x) = 9x, which of the following [#permalink] New post   05 Nov 2020, 12:33
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Sajjad1994 wrote:
If f(x) = (x + 6)^2 and g(x) = 9x, which of the following specifies all the possible values of x for which f(g(x)) < g(f(x)) ?

(A) –2 < x < 2

(B) x > 2

(C) 0 < x < 3

(D) –4 < x < 2

(E) 6 < x < 9

Solution:

We see that f(g(x)) = f(9x) = (9x + 6)^2 and g(f(x)) = g((x + 6)^2) = 9(x + 6)^2. Therefore, we have:

(9x + 6)^2 < 9(x + 6)^2

3^2 * (3x + 2)^2 < 9(x + 6)^2

(3x + 2)^2 < (x + 6)^2

(3x + 2)^2 - (x + 6)^2 < 0

[(3x + 2) - (x + 6)][(3x + 2) + (x + 6)] < 0

[2x - 4][4x + 8] < 0

(x - 2)(x + 2) < 0

x - 2 > 0 and x + 2 < 0 OR x - 2 < 0 and x + 2 > 0

x > 2 and x < -2 OR x < 2 and x > -2

We see that x > 2 and x < -2 is not possible since x can’t be greater than 2 and simultaneously be less than -2. On the other hand, x < 2 and x > -2 is possible, and the two inequalities can be combined to become -2 < x < 2, which is the correct solution set.

Answer: A
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If f(x) = (x + 6)^2 and g(x) = 9x, which of the following [#permalink] New post   05 Nov 2020, 15:13
Expert Reply
Sajjad1994 wrote:
If f(x) = (x + 6)^2 and g(x) = 9x, which of the following specifies all the possible values of x for which f(g(x)) < g(f(x)) ?

(A) –2 < x < 2

(B) x > 2

(C) 0 < x < 3

(D) –4 < x < 2

(E) 6 < x < 9


\(f(g(x)) < g(f(x))\)

\(f(9x) < g((x + 6)^2)\)

\((9x + 6)^2 < 9*(x+6)^2\)

\((3x + 2)^2 < (x + 6)^2\)

\( (3x + 2)^2 - (x + 6)^2 < 0\)

\( (3x + 2 - x - 6)(3x + 2 + x + 6) < 0\)

\((2x -4)(4x + 8) < 0\) in order to have this we need one negative one positive, which means \(2x - 4 > 0\) with \(4x+8 < 0\) or vice versa.

Positive*Negative case: \(x > 2\) and \(x < -2\). No solution.

Negative*Positive case: \(x < 2\) and \(x > -2\), which is \(-2 < x < 2\)


Ans: A
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If f(x) = (x + 6)^2 and g(x) = 9x, which of the following [#permalink]
05 Nov 2020, 15:13
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