AMC 10 · 2010 · #4

Grade 8 algebra
absolute-valueexponentsexponential-functionorder-of-operations sign-analysiscasework ↑ Prerequisites: absolute-value
📏 Long solution 💡 2 insights
Problem
Five expressions are offered and one must be positive for every negative input. Find that expression.

Pick an answer.

(A)
$\frac{x}{\left|x\right|}$
(B)
$-x^2$
(C)
$-2^x$
(D)
$-x^{-1}$
(E)
$\sqrt[3]{x}$
How to solve
Strategy Introduce a Variable

The obvious move is to drop a number like x=-1 into all five expressions. That move is honest for throwing choices away, but it can never crown one: a single value says nothing about the infinitely many other negative numbers. So instead of picking a value of x, I rename it. Every negative x can be written as x=-t with t > 0, and that one substitution rewrites all five expressions as questions about a positive number, where the sign rules are automatic. Before substituting I settle how the expressions are meant to be read, because a minus sign in front of a power is easy to misplace.

1STEP 1

What "must" actually demands

Accepting needs it for every input; rejecting needs just one.

accept P ⇔ P(x) > 0 for all x < 0; reject P ⇔ P(x) ≤ 0 for at least one x < 0
2STEP 2

Read powers before minus signs

Reading powers before minus signs prevents a misreading.

-x²=-(x²), -2^x=-(2^x), -x⁻¹=-(1/x)
3STEP 3

Trade x < 0 for t > 0

Substituting a positive letter makes every sign visible.

x=-t with t > 0; { x : x < 0 }={ -t : t > 0 }
4STEP 4

(A) is always exactly -1

The first candidate is always exactly negative one.

x/|x|=-t/(|-t|)=-t/t=-1 < 0 for all t > 0
5STEP 5

(B): the minus sits outside

The second has its minus outside the square.

-x²=-(-t)²=-t² < 0 for all t > 0
6STEP 6

(C): base 2 stays positive

The third's base keeps it positive before the minus.

-2^x=-2^-t=-1/2^t < 0 for all t > 0
7STEP 7

(E): cube roots keep signs

Cube roots keep the sign of their input.

(∛(x))³=x < 0 ⟹ ∛(x) < 0, ∛(-t)=-∛(t)
8STEP 8

Why (D) is always positive

Only the reciprocal one flips positive, choice (D).

-x⁻¹=-1/x=-1/-t=1/t > 0 for all t > 0
Answer
-x⁻¹
Two points deserve a second look. First, size versus sign: -x⁻¹=1/|x| shrinks as x moves further left, and at x=-100 it is only 0.01. Small is not zero, and 1/t never reaches 0 for a finite t, so "positive" is satisfied. Second, the four rejected expressions came out negative at every x < 0, not just at one convenient value, which is why a quick check at x=-1 happens to point the same way. A different check agrees: at x=-1/4 the values are x/|x|=-1, -x²=-1/16, -2⁻¹/4≈-0.84, ∛(-1/4)≈-0.63, and -x⁻¹=4. Each decoy is a sign argument that stops one step early: (A) hopes |x| repairs the sign, but it only repairs the denominator; (B) hopes the square wins, but the minus is applied after it; (E) borrows the habit that roots are never negative, which holds for square roots and fails for cube roots.
💡Key takeaway

"Must be positive" means positive for every negative x, so rename x as -t with t > 0 one time and read all five signs at once.

  • What "must" actually demands
  • Read powers before minus signs
  • Trade x < 0 for t > 0
  • (A) is always exactly -1
  • (B): the minus sits outside
  • (C): base 2 stays positive
  • (E): cube roots keep signs
  • Why (D) is always positive