GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
f(x) = 3x2 g(x) = x + 1 x is an integer such that –10 ≤ x ≤
[#permalink]
26 Mar 2018, 15:57
26 Mar 2018, 15:57
Expert Reply
4
Bookmarks
00:00
Question Stats:
70% (01:34) correct
29% (01:09) wrong based on 47 sessions
Hide Show timer Statistics
\(f(x) = 3x^2\)
\(g(x) = x + 1\)
x is an integer such that –10 ≤ x ≤ –1.
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
\(g(x) = x + 1\)
x is an integer such that –10 ≤ x ≤ –1.
Quantity A |
Quantity B |
f(g(x)) |
g(f(x)) |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Drill 2
Question: 11
Page: 551
Question: 11
Page: 551
Re: f(x) = 3x2 g(x) = x + 1 x is an integer such that –10 ≤ x ≤
[#permalink]
26 Mar 2018, 17:50
26 Mar 2018, 17:50
1
Answer: A
f (x) = 3 * x^2
g(x) = x + 1
A: f(g(x)) = f(x + 1) = 3 * (x + 1)^2 = 3x^2 + 6x + 3
B: g(f(x) = g (3 * x^2) = 3 * x^2 + 1
So we should compare 6x+3 with 1.
-10 <= x <= -1 so -57 <= 6x+3 <= -3
As 6x+3 is always less than 1, B is bigger than A.
f (x) = 3 * x^2
g(x) = x + 1
A: f(g(x)) = f(x + 1) = 3 * (x + 1)^2 = 3x^2 + 6x + 3
B: g(f(x) = g (3 * x^2) = 3 * x^2 + 1
So we should compare 6x+3 with 1.
-10 <= x <= -1 so -57 <= 6x+3 <= -3
As 6x+3 is always less than 1, B is bigger than A.
Re: f(x) = 3x2 g(x) = x + 1 x is an integer such that –10 ≤ x ≤
[#permalink]
06 Apr 2018, 15:10
06 Apr 2018, 15:10
Expert Reply
Explanation
Plugging 10 values into two compound functions is going to involve lots of arithmetic and will take a long time, so it is better to do this one algebraically.
Working from the inside out, find Quantity A: \(f(g(x))=f(x+1)=3(x+1)^2=3x^2+6x+3\); remember to FOIL the (x + 1) when you square it. Similarly, find Quantity B: \(g(f(x)) = g(3x^2) = 3x^2 + 1\).
You can add or subtract the same value from both quantities without affecting which is bigger; doing so with \(3x^2 + 1\) leaves you with 6x + 2 in Quantity A and 0 in Quantity B.
Because 6x + 2 is a linear function whose graph is a line with positive slope, you know that the values of the function will increase as the values of x increase.
So you only need to plug in the endpoints of the given range of x-values to see what happens to the function: 6(–10) + 2 = –58 and 6(–1) + 2 = –4.
So all possible values of Quantity A are still less than 0, and the answer is choice (B).
Plugging 10 values into two compound functions is going to involve lots of arithmetic and will take a long time, so it is better to do this one algebraically.
Working from the inside out, find Quantity A: \(f(g(x))=f(x+1)=3(x+1)^2=3x^2+6x+3\); remember to FOIL the (x + 1) when you square it. Similarly, find Quantity B: \(g(f(x)) = g(3x^2) = 3x^2 + 1\).
You can add or subtract the same value from both quantities without affecting which is bigger; doing so with \(3x^2 + 1\) leaves you with 6x + 2 in Quantity A and 0 in Quantity B.
Because 6x + 2 is a linear function whose graph is a line with positive slope, you know that the values of the function will increase as the values of x increase.
So you only need to plug in the endpoints of the given range of x-values to see what happens to the function: 6(–10) + 2 = –58 and 6(–1) + 2 = –4.
So all possible values of Quantity A are still less than 0, and the answer is choice (B).
