Question

The real number x when added to its inverse gives the maximum value of the sum at x equal to?

A: 1
B: -1
C: 2
D: -2
THE ANSWER IS (A) FOR YOU REFERENCE

Answer 1

The answer may be C to maximise the value of `x+1/x` over the options given or B identifying a local maximum of the function. The answer could also possibly be D if the sum is wanted rather than `x`.

Explanation:

The word "inverse" in the question is ambiguous, since `x` usually has inverses under both addition and multiplication. More specific terms would be "opposite" (for additive inverse) or "reciprocal" (for multiplicative inverse).

If the question is asking about the additive inverse (opposite), then the sum is always `0` for any `x`. So the sum takes its maximum value for any `x`.

If the question is asking about the multiplicative inverse (reciprocal), then it is asking us to maximise:

`f(x) = x + 1/x`

If `x` is allowed to range over all real numbers then this function has no maximum. Specifically we find that it increases without limit as `x->0^+` and as `x->+oo`.

Possible interpretation 1

Given that this is a multiple choice question, then one interpretation that makes some sense is that we want to choose the option that maximises the value of the function.

We find:

A: `" "f(1) = 1+1/1 = 2`

B: `" "f(-1) = -1+1/(-1) = -2`

C: `" "f(2) = 2+1/2 = 5/2`

D: `" "f(-2) = -2+1/(-2) = -5/2`

So the option that maximises `x+1/x` is C.

Possible interpretation 2

The function `f(x)` has a local maximum when `x=-1`, corresponding to option B.

Here's a graph...

graph{(y-x-1/x)((x+1)^2+(y+2)^2-0.01) = 0 [-10, 10, -5, 5]}

Note that `f(x)` has a local minimum at `x=1` (option A).

Possible interpretation 3

The question might actually be asking for the value of the sum at the maximum rather than the value of `x`. If so, the answer could be D since that is the value of the sum at the local maximum:

`f(-1) = -2`