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Difference between revisions of "1957 AHSME Problems/Problem 29"

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

Latest revision as of 19:53, 20 July 2024

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 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 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
All AHSME Problems and Solutions

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