Help Functions!

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Help Functions!

by tar013 » Mon Jan 14, 2013 2:31 pm
For all non-negative integers x and n such that 0 ≤ x ≤ n, the function fn(x) is defined by the equation fn(x) = xn-x. The smallest value of n for which the maximum of fn(x) occurs when x = 4 is

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

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by GMATGuruNY » Mon Jan 14, 2013 6:48 pm
tar013 wrote:For all non-negative integers x and n such that 0 ≤ x ≤ n, the function f(x,n) is defined by the equation f(x,n) = x^(n-x). The smallest value of n for which the maximum of f(x,n) occurs when x = 4 is

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
The portions in red reflect what I believe is intended.
We can plug in the answers, which represent the smallest value of n for which the maximum of f(x,n) occurs when x = 4.
Since we need the SMALLEST value of n, we should start with the smallest answer choice.

Answer choice A: n=6
When x=0, f(x,n) = x^(n-x) = 0^(6-0) = 0^6 = 0.

When x=1, f(x,n) = x^(n-x) = 1^(6-1) = 1^5 = 1.

When x=2, f(x,n) = x^(n-x) = 2^(6-2) = 2^4 = 16.

When x=3, f(x,n) = x^(n-x) = 3^(6-3) = 3^3 = 27.

When x=4, f(x,n) = x^(n-x) = 4^(6-4) = 4^2 = 16.

When x=5, f(x,n) = x^(n-x) = 5^(6-5) = 5^1 = 5.

When x=6, f(x,n) = x^(n-x) = 6^(6-6) = 6^0 = 1.

The maximum value does not occur when x=4.
Eliminate A.

The values of the function have been highlighted above in red.
They exhibit the following pattern:
The bases are equal to the consecutive integers between 0 and n, inclusive, in ASCENDING order.
When x=4, the base is equal to 4.
The powers are equal to the consecutive integers between n and 0, inclusive, in DESCENDING order.
For the remaining answer choices, we can follow this pattern to determine whether the maximum value occurs when the base is equal to 4.

Answer choice B: n=7
The bases will be 0, 1, 2, 3, 4, 5, 6, 7.
The powers will be 7, 6, 5, 4, 3, 2, 1, 0.
Thus, the values of the function will be:
0�, 1�, 2�, 3�, 4³, 5², 6¹, 7�.
The greatest value here is not 4³.
Eliminate B.

Answer choice C: n=8
The bases will be 0, 1, 2, 3, 4, 5, 6, 7, 8.
The powers will be 8, 7, 6, 5, 4, 3, 2, 1, 0.
Thus, the values of the function will be:
0�, 1�, 2�, 3�, 4�, 5³, 6², 7¹, 8�.
The greatest value here is 4�.

The correct answer is C.
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by tar013 » Mon Jan 14, 2013 8:08 pm
Thank you Mitch. That is indeed the correct answer and I understand your approach. The question was written Fn(x)=x^(n-x) (n as a subscript before the (x) = sign. Can you help me clarify its significance and how its meaning compares to f(x,n)? I don't know what it is about function problems that give me such a hard time. Also, is there an algebraic way to solve this? Thanks again.

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by pemdas » Tue Feb 19, 2013 8:37 am
GMATGuruNY wrote:
tar013 wrote:For all non-negative integers x and n such that 0 ≤ x ≤ n, the function f(x,n) is defined by the equation f(x,n) = x^(n-x). The smallest value of n for which the maximum of f(x,n) occurs when x = 4 is

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
f(x,n)=x^(n-x)
d/dx = -x^(n-x-1)*(-n+x+x ln(x)), for x=4 we get -4^(n-5)*(-n+4+4 ln(4))=0. Cancell out -4^(n-5) and leave -n+4+4 ln(4)=0 to solve for min (n), -n+8 ln(4)=0. Cancell out ln(4)to solve -n+8=0 and it appears c is the answer, because n=8.
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by Soumita Ghosh » Thu Feb 21, 2013 11:28 pm
can anyone show me alternate approach for this problem.

@Mitch I am not understanding why you are taking X=0,1,2,3,4,5,6??It is given in the problem that we need to take x=4 in order to get the maximum of f(x,n).We need to find the smallest value of n keeping in mind that x=4 and f(x,n) must be maximum.


I am understanding anything wrong??

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by [email protected] » Fri Feb 22, 2013 1:41 am
Soumita Ghosh wrote:...It is given in the problem that we need to take x=4 in order to get the maximum of f(x,n).We need to find the smallest value of n keeping in mind that x=4 and f(x,n) must be maximum.

I am understanding anything wrong??
I think the simple format for representing the function introduced by Mitch is creating a bit confusion here. f_n(x) i.e. 'f subscript n' as the image below is essentially different from f(x, n).

Image

f_n(x) means the the primary variable of the function is x, n is a parameter of the function. Whereas, f(x, n) means x and n both are variables of the function.

Without going into much details about variables and parameters, just know that f_n(x) is a family of functions, where x is the variable and n determines the nature of the function.

For different values of n, we'll have different f(x) as f(x) = x^(n - x)
Hence,
  • For n = 1, f(x) = x^(1 - x) where 0 ≤ x ≤ 1
    For n = 2, f(x) = x^(2 - x) where 0 ≤ x ≤ 2
    For n = 3, f(x) = x^(3 - x) where 0 ≤ x ≤ 3
    ...
    For n = 10, f(x) = x^(10 - x) where 0 ≤ x ≤ 10 etc
Now, the question is asking for the smallest value of n for which f(x) is maximum when x = 4.

This means you have to first fix n, then look if f(x) is attaining the maximum value at x = 4 or not. It is possible that for more than one value of n, f(x) will attain maximum value at x = 4. Hence, they are asking for the smallest possible value of n.

If you fix x first, i.e. take x = 4 as your starting point, you are basically converting a function of x to a function of n, i.e. you are making f_n(4) = 4^(n - 4). Now, it is not a function at all.

Also 4^(n - 4) will keep on increasing with increasing values of n. So the question (the smallest value of n for which the function attains maximum value when x = 4) doesn't make any sense.

Soumita Ghosh wrote:can anyone show me alternate approach for this problem.
I'm afraid that only two methods to solve this problem is already posted. For this kind of problems Calculus is the way to go if you are looking for a less time consuming method. Mitch's method is simpler and for those who do not know Calculus.

However, if you try to visualize the function, you'll see (base + power) is always equal to n and the function values always starts with 0^n and then the base keeps on increasing and the power keeps on decreasing, and finally finishes with n^0. Hence, the value of the function starts from 0 and then keeps on increasing to attain a maximum value and then decreases to 1.

Hence, a good point to start checking for maximum value of the function is where power = base, i.e. x = n/2.
In this case, x = 4.
Hence, let's start with n = 2*x = 8

f(4) = 4^(8 - 4) = 4^4 = 256
Then check the values of f(3) and f(5) to verify whether f(4) is the maximum or not.
f(3) = 3^(8 - 3) = 3^5 = 243
f(5) = 5^(8 - 3) = 5^3 = 125

Hence, for n = 8, the function attains maximum value at x = 4
So we don't need to check for any value of n > 8

Now, let's check for n = 7 ---> n/2 = 3.5
So we'll check for integer values of x on both side of 3.5, i.e. x = 3 and x = 4
f(3) = 3^(7 - 3) = 3^4 = 81
f(4) = 4^(7 - 3) = 4^3 = 64 < f(3)

No need to check for the values of n < 7, as evidently 4 will be far away from the center point, i.e. x = n/2 and f(4) will be smaller than the maximum value.
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by GMATGuruNY » Fri Feb 22, 2013 7:31 am
Soumita Ghosh wrote:can anyone show me alternate approach for this problem.

@Mitch I am not understanding why you are taking X=0,1,2,3,4,5,6??It is given in the problem that we need to take x=4 in order to get the maximum of f(x,n).We need to find the smallest value of n keeping in mind that x=4 and f(x,n) must be maximum.

I am understanding anything wrong??
The intent of original problem is that the operation x^(n-x) is to be performed on ALL INTEGER VALUES OF X BETWEEN 0 AND N, iNCLUSIVE.
Thus, when n=7, the operation x^(7-x) must be performed on all integer values of x between 0 and 7, inclusive, as shown in my solution above.
The correct answer choice is the smallest value of n such that -- when the operation x^(n-x) is performed on all of the appropriate values of x -- x=4 yields the greatest result.

The original problem likely uses the notation kindly supplied by Anurag:
Image
This notation is beyond the scope of the GMAT.
In fact, this entire problem is beyond the scope of the GMAT.
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