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devilmirror
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For any real number x and y, |x + y| <= |x| + |y| |x^3 30 Nov 2007, 16:51
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For any real number x and y, |x + y| <= |x| + |y|

|x^3 - 2x^2 + 3x - 4| <= M. What is the maximum value of M when -3 <= x <= 2?

A) 4
B) -4
C) 2
D) 58
E) 64



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For any real number x and y, |x + y| <= |x| + |y| |x^3 30 Nov 2007, 20:25
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D.

|x^3 - 2x^2 + 3x - 4| <= M. (-3 <= x <= 2)

1.x<0> min at x=-3
p=-27-18-9-4=-58

2. x>0 p=x^3 - 2x^2+ 3x - 4<=x^3 + 3x<=14

M=max(58,14)=58
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For any real number x and y, |x + y| <= |x| + |y| |x^3 01 Dec 2007, 12:25
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D.

|x^3 - 2x^2 + 3x - 4| <= M. (-3 <= x <= 2)

1.x<0> min at x=-3
p=-27-18-9-4=-58

2. x>0 p=x^3 - 2x^2+ 3x - 4<=x^3 + 3x<=14

M=max(58,14)=58
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LOL, how should I say this? Your answer is right but it is wrong. If the question changed +3x to -3x, you would get the wrong answer.

Solution:

Since |x + y| <= |x| + |y|
|x^3 - 2x^2 + 3x - 4| <= |x^3| + |-2x^2| + |3x| + |-4|
<==> |x|^3 + 2|x|^2 + 3|x| + 4 ...(1)

And since -3 <= x <= 2
0 <= |x| <= 3
The maximum value for |x| is 3. Use this number in (1)
M = 58
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For any real number x and y, |x + y| <= |x| + |y| |x^3 01 Dec 2007, 12:41
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let's try 58. It's ok. 64 is impossible, so the OA is D.
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For any real number x and y, |x + y| <= |x| + |y| |x^3 01 Dec 2007, 12:58
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I really do not like |x + y| <= |x| + |y|

for example:
|x^2 +2x -15| <= M. What is the maximum value of M when -2 <= x <= 4?

1. |x^2 +2x -15|<=|x^2|+|2x|+|-15| ==> for x=4: 16+8+15=39

2. |x^2 +2x -15| for x=-2: |4-4-15|=15
for x=4: |16+8-15|=9 ==> max(15,9)=15 more precise but also incorrect

3. right answer. x=-1 |1-2-15|=16 so |x^2 +2x -15|<=16

the most common approach (maybe out of GMAT)
f(x)=x^2 +2x -15

max(|f(-2 <= x <= 4)|)=max(|f(-2)|,|f(4)|,|f(p)|) where p - a point of extreme (extrema)

f'(x)=2x+2=0 ==>2p+2=0 ==> p=-1

so max(|f(-2)|,|f(4)|,|f(-1)|)=max(15,9,16)=16
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For any real number x and y, |x + y| <= |x| + |y| |x^3 01 Dec 2007, 13:36
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walker
I really do not like |x + y| <= |x| + |y|

for example:
|x^2 +2x -15| <= M. What is the maximum value of M when -2 <= x <= 4?

1. |x^2 +2x -15|<=|x^2|+|2x|+|-15| ==> for x=4: 16+8+15=39

2. |x^2 +2x -15| for x=-2: |4-4-15|=15
for x=4: |16+8-15|=9 ==> max(15,9)=15 more precise but also incorrect

3. right answer. x=-1 |1-2-15|=16 so |x^2 +2x -15|<=16

the most common approach (maybe out of GMAT)
f(x)=x^2 +2x -15

max(|f(-2 <= x <= 4)|)=max(|f(-2)|,|f(4)|,|f(p)|) where p - a point of extreme (extrema)

f'(x)=2x+2=0 ==>2p+2=0 ==> p=-1

so max(|f(-2)|,|f(4)|,|f(-1)|)=max(15,9,16)=16
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Yes, that's the usual approach by derivating and identifying the extremums at f'(x) = 0.

I'm also directed to use this one :)
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For any real number x and y, |x + y| <= |x| + |y| |x^3 01 Dec 2007, 15:27
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walker
I really do not like |x + y| <= |x| + |y|

for example:
|x^2 +2x -15| <= M. What is the maximum value of M when -2 <= x <= 4?

1. |x^2 +2x -15|<=|x^2|+|2x|+|-15| ==> for x=4: 16+8+15=39

2. |x^2 +2x -15| for x=-2: |4-4-15|=15
for x=4: |16+8-15|=9 ==> max(15,9)=15 more precise but also incorrect

3. right answer. x=-1 |1-2-15|=16 so |x^2 +2x -15|<=16

the most common approach (maybe out of GMAT)
f(x)=x^2 +2x -15

max(|f(-2 <= x <= 4)|)=max(|f(-2)|,|f(4)|,|f(p)|) where p - a point of extreme (extrema)

f'(x)=2x+2=0 ==>2p+2=0 ==> p=-1

so max(|f(-2)|,|f(4)|,|f(-1)|)=max(15,9,16)=16
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Maybe my question is not 100% clear.

However, the question does not ask for the maximum point on graph. It ask what is the possible maximum value M in this inequation. The method that you are using is for solving equation.

Therefore, using |x + y| <= |x| + |y|

|x^2 +2x -15|<=|x^2|+|2x|+|-15| ==> for x=4: 16+8+15=39

M is still 39 not 16.

Any opinion?
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For any real number x and y, |x + y| <= |x| + |y| |x^3 01 Dec 2007, 21:14
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devilmirror
However, the question does not ask for the maximum point on graph. It ask what is the possible maximum value M in this inequation. The method that you are using is for solving equation.

Therefore, using |x + y| for x=4: 16+8+15=39

M is still 39 not 16.

Any opinion?
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You are right if GMAT has cases like (3,7,39,100,299).
But in the cases (8,15,16,39,42) your answer is not good.



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