For which of the following functions does f(x)=f(x) for all real num

Question

For which of the following functions does f(−x)=−f(x) for all real number values of x?

A. f(x) = x^8−x^4
B. f(x) = x^2−x^6
C. f(x) = x^5/x^7
D. f(x) = x^9/(x^5+1)
E. f(x) = x^5/(x^2+1)

kdatt1991 · Accepted Answer

A. f(-x) = (-x)^8 - (-x)^4 is not equal to -f(x) = -(x^8 - x^4). B. option B has a similar structure to option A's answer choice!− f(-x) = (-x)^2 - (-x)^6 which is not equal to -f(x) = -(x^2 - x^6). C. f(-x) = (-x)^5/(-x)^7 is also not equal to -f(x) = -(x^5/x^7). f(-x) yields a + answer, whilst -f(x) yields a − answer. D. f(-x) = (-x)^9/((-x)^5+1) is not equal as -f(x) = (-1)*x^9/(x^5+1). E. f(-x) = (-x)^5/((-x)^2+1) is equal to -f(x) = (-1)*x^5/(x^2+1), because (-x)^5 is equal to (-1)*x^5 and ((-x)^2+1) is the same as (x^2+1). Please consider giving me KUDOS if you felt this post was helpful! Thanks.

PareshGmat · Answer

Answer = E. x^5/(x^2+1)

Change of sign happens does NOT happen when:

1. All powers are even

2. All powers are odd

If you observe the OA, only option E has the combination of odd power in numerator & even in denominator

f(-x) = \frac{(-x)^5}{(-x)^2+1} = \frac{-x^5}{x^2+1}

-f(x) = \frac{-x^5}{x^2+1}

Answer = E

Cadaver · Answer

E = -X^5/(X^2+1)...Both the sides

gmat6nplus1 · Answer

If f(-x)=-f(x) then f(-x) + f(x) = 0

E) f(-x)=  -x^5 /x^2+1; f(x) is option E itself. Same denominator, subtract numerator you get 0.

Answer E.

sauberheine1 · Answer

Use x = 2 as your guide to test the answer choices.

For choice  A , f(−x) will be the same as f(x), because the even exponents ensure that whether you're taking 2 or -2 to the 8th or 4th powers the results will always be the same. Therefore, since f(−x)=f(x), then this doesn't satisfy the prompt "does f(−x)=−f(x).

Mathematically, using x = 2 you'll have f(−x)=(−2)8−(−2)4=256−16=230. Meanwhile, −f(x)=−(28−24)=−(256−16)=−230, so you don't have a match.

For choice  B , the logic and math are pretty similar. Since both exponents are even, whether you're using a positive or negative value as your input, the results will be the same either way for f(2) and f(−2). So your positive/negative eyeball test should tell you that this one won't work either.

Mathematically, using x = 2, you'll have: f(−x)=(−2)2−(−2)6=4−64=−60. And −f(x)=−(22−26)=−(4−64)=−(−60)=60 They're not the same, so choice B doesn't hold.

Choice  C  is similar, in that x^5/x^7 is going to net out to 1x/^2. A negative input like 2 will still yield a positive output (1/4), whereas −f(x) is going to yield a negative number (f(x)=1/4, so −f(x)=−1/4).

By now you should see that choice  D  will be the same - with two odd exponents, the negative input will yield a negative divided by a negative, and a positive number overall. While taking −f(x) will yield a negative.

Mathematically, that's f(−x)=(−2)^9/(−2)^5+1=−29−31, which is a positive number, whereas −f(x)=−(2925+1), where the answer will clearly be negative.

Only choice  E  mixes an odd exponent with an even exponent, a necessary pairing to satisfy the question. Mathematically with 2, you'd have:

f(−x)=(−2)^5/(−2^)2+1=−32/5, and

−f(x)=−2^5/2^2+1=−32/5, for a match.

ANSWER: E

sm1510 · Answer

A.  f(-x) = (-x)^8 - (-x)^4  is not equal to  -f(x) = -(x^8 - x^4).

B. option B is the same as option A−  f(-x) = (-x)^2 - (-x)^6  is not equal to  -f(x) = -(x^2 - x^6).

C.  f(-x) = \frac{(-x)^5}{(-x)^7}  is not equal to  -f(x) = \frac{-(x^5)}{(x^7)}  . f(-x)  yields a + answer, whilst  -f(x)  yields a − answer.

D.  f(-x) = \frac{(-x)^9}{((-x)^5+1)}  is not equal as  -f(x) = \frac{(-1)*x^9}{(x^5+1)} .

E.  f(-x) = \frac{(-x)^5}{((-x)^2+1)}  is equal to  -f(x) = \frac{(-1)*x^5}{(x^2+1)} , because  (-x)^5  is equal to  (-1)*x^5  and  ((-x)^2+1)  is the same as  (x^2+1) .

Bambi2021 · Answer

Let x = 1

A: f(-1) -> 0
B: f(-1) -> 0
C: f(-1) -> 1, and f(1) = 1
D: f(-1) -> -1/0 
E: f(-1) -> -1/2, and f(1) = 1/2

Answer: E

Alvaro20172027 · Answer

I misread the answer D as if the +1 were adding the exponent, in which case it would become x^6 and D would be the answer.

onlymalapink · Answer

none of the solutions make any sense no explanation on why e works but d doesnt makes sense to me?

Bunuel · Answer

Note that (-x)^5 = -x^5, while (-x)^2 = x^2.
_______________________________________

For E, if  f(x) = \frac{x^5}{x^2+1} , then:

f(-x) =\frac{ (-x)^5}{(-x)^2+1}=\frac{ -x^5}{x^2+1}=- \frac{x^5}{x^2+1}

and

-f(x) =-\frac{x^5}{x^2+1}

As you can see, these two expressions match.
_______________________________________

For D, if  f(x) = \frac{x^9}{x^5+1} , then:

f(-x) =\frac{ (-x)^9}{(-x)^5+1}=\frac{ -x^9}{-x^5+1}= -\frac{ x^9}{-x^5+1}

and

-f(x) =-\frac{x^9}{x^5+1}

As you can see, these two expressions do not match.
_______________________________________

Hope it's clear now.

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For which of the following functions does f(x)=f(x) for all real num

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