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IEsailor
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If x is a number such that –2 ≤ x ≤ 2 which of the following Updated on: 02 Jul 2013, 10:05
Originally posted by IEsailor on 10 Apr 2011, 21:20.
Last edited by Bunuel on 02 Jul 2013, 10:05, edited 2 times in total.
Renamed the topic and edited the question.
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If x is a number such that –2 ≤ x ≤ 2, which of the following has the largest possible absolute value?

A. 3x – 1
B. x^2 + 1
C. 3 – x
D. x – 3
E. x^2 – x
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A
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If x is a number such that –2 ≤ x ≤ 2 which of the following 21 Jun 2017, 15:53
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x is a number such that –2 ≤ x ≤ 2

In order to solve this problem you can substitute values of +2 and -2 based and can quickly get the absolute value.

A. 3x – 1 =====> 3*-2-1 = -7 = l-7l = 7 ======> Maximum ab. value
B. x^2 + 1 =====> 4+1 = 5
C. 3 – x =====> 2-(-2) = 4
D. x – 3 =====> 3-(-2) = 5
E. x^2 – x =====> 4-(-2) = 6

So, only 7 is the maximum value.

A is the correct answer.
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If x is a number such that –2 ≤ x ≤ 2 which of the following 11 Apr 2011, 14:17
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IEsailor
If x is a number such that –2 ≤ x ≤ 2, which of the following has the largest possible absolute value?

3x – 1
x2 + 1
3 – x
x – 3
x2– x

Can this be done Algebrically / graphically.

I got the answer using substitution but found that time consuming.
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If you know how to plot graphs, this is very straight forward, most of the expressions are very simple to plot.

In case you want to do algebraically, here are a couple of simple rules you need to know:
(a) For linear expressions, the maxima or minima will always be at the extreme ends of a range
(b) For quadratic expressions, the maxima or minima will always either be the extreme end of a range of the global max/min of the function


3x – 1 : At ends, value is 5 and -7
x2 + 1 : At ends, value is 5 and global min value is at x=0, where it is 1
3 – x : At ends, value is 1, 5
x – 3 : At ends, value is -5,-1
x2– x : At ends, value is 2,6. Global min value is at x=0.5, where it is -0.25

(Its easy to tell what the global min for these simple quadratic functions is knowing where there roots are, it is half way between the roots)

Clearly, answer is A
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If x is a number such that –2 ≤ x ≤ 2 which of the following 11 Apr 2011, 23:40
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hello shrouded1
I don't understand your last statement. Pls can you elaborate.

(Its easy to tell what the global min for these simple quadratic functions is knowing where there roots are, it is half way between the roots)

On way to get the global min is applying the calculus. -
y = x^2+1
dy/dx = 0 or 2x = 0. So the min occurs at x=0

Similarly y = x^2-x
dy/dx = 0 or 2x - 1 = 0 . So the min occurs at x = 1/2 = 0.5
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If x is a number such that –2 ≤ x ≤ 2 which of the following 11 Apr 2011, 23:47
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gmat1220
hello shrouded1
I don't understand your last statement. Pls can you elaborate.

(Its easy to tell what the global min for these simple quadratic functions is knowing where there roots are, it is half way between the roots)

On way to get the global min is applying the calculus. -
y = x^2+1
dy/dx = 0 or 2x = 0. So the min occurs at x=0

Similarly y = x^2-x
dy/dx = 0 or 2x - 1 = 0 . So the min occurs at x = 1/2 = 0.5
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Thats implicitly what I did. But I used a little trick so I dont have to do the calculations.
For any quadratic function with real roots, the minima/maxima (depending on wether the coefficient of x^2 is positive or negative) will occur at the mean of the roots. (This is very easy to prove, all I am saying is that the minima is -b/2a, which you can show by differentiation).

So for x^2-x = x(x-1) the roots are 0,1 and the minima will be at 0.5

For x^2+1, the roots are not real, but this function is simply a vertically translated version of x^2=0, hence it also minimizes/maximizes at the same level which is x=0.



Generalizing, you can always use the fornula x=(-b/2a) for the min or max valuation
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If x is a number such that –2 ≤ x ≤ 2 which of the following 02 Jul 2013, 10:37
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If x is a number such that –2 ≤ x ≤ 2, which of the following has the largest possible absolute value?

A. 3x – 1
B. x^2 + 1
C. 3 – x
D. x – 3
E. x^2 – x

First of all notice that we can eliminate options C and D right away: \(|3-x|=|x-3|\), so these two options will have the same maximum value and since we cannot have two correct answers in PS questions then none of them is correct.

Evaluate each option by plugging min and max possible values of x:

A. 3x – 1 --> max for x=-2 --> |3*(-2)-1|=7.


B. x^2 + 1 --> max for x=-2 or x=2 --> |2^2 + 1|=5.


E. x^2 – x --> max for x=-2 --> |(-2)^2 - (-2)|=6.


Answer: A.

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