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If f(x) = 4x^2 - 36x + 77 for all real values of x from -10 and 10, 09 Jul 2006, 02:49
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If f(x) = 4x^2 - 36x + 77 for all real values of x from -10 and 10, then the range of all possible values of f is

(A) 537
(B) 539
(C) 541
(D) 543
(E) none of these
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E
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If f(x) = 4x^2 - 36x + 77 for all real values of x from -10 and 10, 24 Jul 2015, 01:30
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riyazgilani


How to decide on min and max values for a function..?
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Given a quadratic, say \(f(x) = ax^2 + bx + c\), it will either have a maximum value or a minimum value.
If it is an upward open parabola (happens when 'a' is positive) as shown in the figure in the post above, it will have a minimum value. It's maximum value will be infinity.
On the flip side, if it is a downward open parabola (happens when 'a' is negative), it will have a maximum value. Its minimum value will be - infinity.

This maximum or minimum value (as the case may be) will be found at x = -b/2a

Here \(f(x) = 4x^2 -36x+77\)
a = 4 which is positive.
So it is an upward opening parabola and will have a minimum value when \(x = -b/2a = -(-36)/2*4 = 9/2\)
In this case, \(f(9/2) = 4*(9/2)^2 - 36*(9/2) + 77 = -4\)

The maximum value for this parabola is infinity. But we are given that the range of x is from -10 to 10. So the parabola will take maximum value when x is as far away from 9/2 as possible. -10 is the farthest from 9/2 so the function will take maximum value at x = -10

\(f(-10) = 4(-10)^2 -36(-10)+77 = 837\)

Range = 837 - (-4) = 841
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If f(x) = 4x^2 - 36x + 77 for all real values of x from -10 and 10, 09 Jul 2006, 07:06
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kevincan
If f(x)= 4x^2-36x+77 for all real values of x from -10 and 10, then the range of all possible values of f is

(A) 537 (B) 539 (C) 541 (D) 543 (E) none of these
Show more



f(x)= 4x^2-36x+77

f(x) has highest value when x = -10.
f(x= -10) = 400+360+77 = 837

f(x) has lowest value value when x = 8
f(x= 8) = 4 (8x8) - 36 (8) +77 = 8 (32-36) + 77 = - 45.

the range = 837 - (-45) = 882.

SO E.
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If f(x) = 4x^2 - 36x + 77 for all real values of x from -10 and 10, 09 Jul 2006, 07:22
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kevincan
If f(x)= 4x^2-36x+77 for all real values of x from -10 and 10, then the range of all possible values of f is

(A) 537 (B) 539 (C) 541 (D) 543 (E) none of these
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f(x)= 4x^2 -36x+77 = (2x-9)^2 - 9^2 + 77= (2x-9)^2 - 4 >= -4 because (2x-9)^2 >= 0
The equality happens when 2x-9= 0 --> x= 9/2 ( still in the range from -10 to 10) ---> minimum value of f(x)= -4
The highest value is when x=-10 as Himalaya pointed out.
---> the range = 837 - ( -4) = 841 --> E it is.
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If f(x) = 4x^2 - 36x + 77 for all real values of x from -10 and 10, 09 Jul 2006, 07:34
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Quote:
f(x)= 4x^2-36x+77

f(x) has highest value when x = -10.
f(x= -10) = 400+360+77 = 837

f(x) has lowest value value when x = 8
f(x= 8) = 4 (8x8) - 36 (8) +77 = 8 (32-36) + 77 = 45.
the range = 837 - (-45) = 882.

SO E.
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oh i didnot consider fractions. i know nothing can escaped from you in qt. you are :king
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If f(x) = 4x^2 - 36x + 77 for all real values of x from -10 and 10, 09 Jul 2006, 07:42
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hik, i'm not a quant queen, still need to improve myself a lot ;)

Hey, buddy, i notice those Nepalese songs in your signature, lemme enjoy them ;)
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If f(x) = 4x^2 - 36x + 77 for all real values of x from -10 and 10, 09 Jul 2006, 08:05
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hik, i'm not a quant queen, still need to improve myself a lot ;)

Hey, buddy, i notice those Nepalese songs in your signature, lemme enjoy them ;)
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undoubtly its my gr8 pleasure. :king
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If f(x) = 4x^2 - 36x + 77 for all real values of x from -10 and 10, 23 Jul 2015, 03:54
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laxieqv
kevincan
If f(x)= 4x^2-36x+77 for all real values of x from -10 and 10, then the range of all possible values of f is

(A) 537 (B) 539 (C) 541 (D) 543 (E) none of these

f(x)= 4x^2 -36x+77 = (2x-9)^2 - 9^2 + 77= (2x-9)^2 - 4 >= -4 because (2x-9)^2 >= 0
The equality happens when 2x-9= 0 --> x= 9/2 ( still in the range from -10 to 10) ---> minimum value of f(x)= -4
The highest value is when x=-10 as Himalaya pointed out.
---> the range = 837 - ( -4) = 841 --> E it is.
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How to decide on min and max values for a function..?
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If f(x) = 4x^2 - 36x + 77 for all real values of x from -10 and 10, 23 Jul 2015, 05:05
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riyazgilani
laxieqv
kevincan
If f(x)= 4x^2-36x+77 for all real values of x from -10 and 10, then the range of all possible values of f is

(A) 537 (B) 539 (C) 541 (D) 543 (E) none of these

f(x)= 4x^2 -36x+77 = (2x-9)^2 - 9^2 + 77= (2x-9)^2 - 4 >= -4 because (2x-9)^2 >= 0
The equality happens when 2x-9= 0 --> x= 9/2 ( still in the range from -10 to 10) ---> minimum value of f(x)= -4
The highest value is when x=-10 as Himalaya pointed out.
---> the range = 837 - ( -4) = 841 --> E it is.

How to decide on min and max values for a function..?
Show more

For all GMAT questions that ask for min./max. values, you will be able to create squares of the form (a+b)^2 or (a-b)^2. This way you can minimise the value by putting a=b for (a-b)^2. This is true as a square can be \(\geq 0\) with the minimum value of a square = 0.

Similarly, for this question, 4x^2 -36x+77 = (2x-9)^2 - 9^2 + 77= (2x-9)^2 - 4 and this will be minimum when (2x-9) = 0---> x = 4.5

You can also look at it graphically. The given equation : 4x^2 -36x+77 represents a parabola (general equation : ax^2+bx+c=0) as shown in the attached picture.

So the minimum value will occur at x =4.5 and as the shape of the parabola is opening upwards.

In the range \(-10\leq x \leq 10\) , we see that the maximum will occur at x =-10 (as b <0 )

Either way, you will get the same answer.
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Parabola.jpg [ 25.3 KiB | Viewed 5731 times ]

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If f(x) = 4x^2 - 36x + 77 for all real values of x from -10 and 10, 14 Jan 2026, 13:10
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is this really a sub 505 level question??
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