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f(x) = 3x2 g(x) = x + 1 x is an integer such that –10 ≤ x ≤
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26 Mar 2018, 15:57
26 Mar 2018, 15:57
Expert Reply
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69% (01:35) correct
30% (01:09) wrong based on 46 sessions
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\(f(x) = 3x^2\)
\(g(x) = x + 1\)
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 ≤
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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 ≤
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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).
