# 1957 AHSME Problems/Problem 29

## Problem

The relation $x^2(x^2 - 1)\ge 0$ is true only for:

$\textbf{(A)}\ x \ge 1\qquad \textbf{(B)}\ - 1 \le x \le 1\qquad \textbf{(C)}\ x = 0,\, x = 1,\, x = - 1\qquad \\\textbf{(D)}\ x = 0,\, x\le-1,\, x\ge 1\qquad\textbf{(E)}\ x\ge 0$

## Solution

To solve this problem, think about the graph of $f(x)=x^2(x^2-1)=x^2(x-1)(x+1)$. The function equals zero only at the values $-1$, $0$, and $1$. Because the function is a quartic polynomial with a positive leading coefficient, it will go to positive infinity as $x$ tends to either positive or negative infinity. Thus, when $x$ is greatly negative, the function will be positive, and so the given inequality will hold. As $x$ gradually becomes more positive, it will eventually equal $-1$, when the function will equal zero. Thus, the $(x+1)$ term will switch to being positive, and so the whole function will become negative, where the inequality does not hold. $f(x)$ again reaches $0$ when $x=0$, but here the $x^2$ term does not change sign, so the function stays negative afterwards. Finally, when $x=1$, the function crosses the $x$-axis as the $(x-1)$ term changes sign, and the function goes off to positive infinity. Thus, the function is positive (and thus the given inequality will hold) when $x \leq -1$, $x=0$, and $x \geq 1$, which is answer choice $\fbox{\textbf{(D)}}$.

## See Also

 1957 AHSC (Problems • Answer Key • Resources) Preceded byProblem 28 Followed byProblem 30 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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