document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209860>Question 1209860</A>: <I><SPAN STYLE=\"word-break: break-all;\"> For each of the following functions, determine if the function is increasing, decreasing, even, odd, and/or invertible on its natural domain. Select all the properties that apply.<BR>\n" );
document.write( "f(x) = \frac{\sqrt{|x|}}{x} + x + \frac{1}{x^2}<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850884 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850884\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's analyze the function f(x) = (√|x| / x) + x + (1/x²) to determine its properties.\r<BR>\n" );
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document.write( "**1. Natural Domain:**\r<BR>\n" );
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document.write( "* The square root requires |x| ≥ 0, which is true for all x.<BR>\n" );
document.write( "* The terms 1/x and 1/x² require x ≠ 0.<BR>\n" );
document.write( "* Therefore, the natural domain is all real numbers except 0, or (-∞, 0) U (0, ∞).\r<BR>\n" );
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document.write( "**2. Even or Odd:**\r<BR>\n" );
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document.write( "* **Even:** f(-x) = f(x)<BR>\n" );
document.write( "* **Odd:** f(-x) = -f(x)\r<BR>\n" );
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document.write( "Let's test f(-x):\r<BR>\n" );
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document.write( "f(-x) = (√|-x| / -x) - x + (1/(-x)²)<BR>\n" );
document.write( "f(-x) = (√|x| / -x) - x + (1/x²)<BR>\n" );
document.write( "f(-x) = -(√|x| / x) - x + (1/x²)\r<BR>\n" );
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document.write( "Now, let's see if f(-x) = f(x) or f(-x) = -f(x):\r<BR>\n" );
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document.write( "* f(x) = (√|x| / x) + x + (1/x²)<BR>\n" );
document.write( "* -f(x) = -(√|x| / x) - x - (1/x²)\r<BR>\n" );
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document.write( "Comparing:\r<BR>\n" );
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document.write( "* f(-x) ≠ f(x) (Not even)<BR>\n" );
document.write( "* f(-x) ≠ -f(x) (Not odd)\r<BR>\n" );
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document.write( "Therefore, the function is **neither even nor odd**.\r<BR>\n" );
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document.write( "**3. Increasing or Decreasing:**\r<BR>\n" );
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document.write( "* To determine if the function is increasing or decreasing, we need to analyze its derivative, f'(x).\r<BR>\n" );
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document.write( "Let's break down f(x) into cases:\r<BR>\n" );
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document.write( "* **For x > 0:** f(x) = (√x / x) + x + (1/x²) = x^(-1/2) + x + x^(-2)<BR>\n" );
document.write( "* **For x < 0:** f(x) = (√(-x) / x) + x + (1/x²) = -(-x)^(-1/2) + x + x^(-2)\r<BR>\n" );
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document.write( "Now, find the derivatives:\r<BR>\n" );
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document.write( "* **For x > 0:**<BR>\n" );
document.write( "    * f'(x) = (-1/2)x^(-3/2) + 1 - 2x^(-3) = (-1/2x√x) + 1 - (2/x³)<BR>\n" );
document.write( "* **For x < 0:**<BR>\n" );
document.write( "    * f'(x) = (-1/2)(-x)^(-3/2) + 1 - 2x^(-3) = (1/2(-x)√-x) + 1 - (2/x³)\r<BR>\n" );
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document.write( "Analyzing the derivatives is complex and doesn't reveal a simple answer. It's difficult to make a general statement about increasing or decreasing behavior across the entire domain.<BR>\n" );
document.write( "Therefore, we cannot easily state if the function is increasing or decreasing. A graphing calculator would be helpful to determine this.\r<BR>\n" );
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document.write( "**4. Invertibility:**\r<BR>\n" );
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document.write( "* A function is invertible if it is one-to-one (passes the horizontal line test).<BR>\n" );
document.write( "* Since the function is not strictly increasing or decreasing across its entire domain, it is **not invertible**. Also, because it is not even or odd, it is likely not invertible.\r<BR>\n" );
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document.write( "**Conclusion:**\r<BR>\n" );
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document.write( "* **Domain:** (-∞, 0) U (0, ∞)<BR>\n" );
document.write( "* **Neither even nor odd**<BR>\n" );
document.write( "* **Invertible:** No<BR>\n" );
document.write( "* **Increasing/Decreasing:** Difficult to determine without further analysis (graphing is recommended).<BR>\n" );
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