document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1164420>Question 1164420</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let r>1 be a rational number. Prove that for any x->[0,1],\r<BR>\n" );
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document.write( "1/2^(r-1) <= x^(r) + (1-x)^(r) <= 1 \r<BR>\n" );
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document.write( "**Note: I am supposed to use differentiation to find the gradient of an equation in the inequality. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #854349 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2264)</B></A><img src=\"/images/stars/stars-13.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=854349\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To prove the inequality $\frac{1}{2^{r-1}} \le x^r + (1-x)^r \le 1$ for $x \in [0, 1]$ and $r > 1$, we define a function $f(x)$ and analyze its behavior using differentiation.\r<BR>\n" );
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document.write( "### 1. Define the Function<BR>\n" );
document.write( "Let $f(x) = x^r + (1-x)^r$ for $x \in [0, 1]$. Since $r > 1$, $f(x)$ is continuous and differentiable on this interval.\r<BR>\n" );
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document.write( "### 2. Find the Gradient (Derivative)<BR>\n" );
document.write( "To find the critical points, we take the derivative of $f(x)$ with respect to $x$:<BR>\n" );
document.write( "$$f'(x) = rx^{r-1} + r(1-x)^{r-1}(-1)$$<BR>\n" );
document.write( "$$f'(x) = r\left[ x^{r-1} - (1-x)^{r-1} \right]$$\r<BR>\n" );
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document.write( "### 3. Identify Critical Points<BR>\n" );
document.write( "Set the gradient to zero to find the extrema:<BR>\n" );
document.write( "$$r\left[ x^{r-1} - (1-x)^{r-1} \right] = 0$$<BR>\n" );
document.write( "Since $r > 1$, we divide by $r$:<BR>\n" );
document.write( "$$x^{r-1} = (1-x)^{r-1}$$<BR>\n" );
document.write( "Taking the $(r-1)$-th root of both sides (which is valid as $x \ge 0$ and $1-x \ge 0$):<BR>\n" );
document.write( "$$x = 1 - x \implies 2x = 1 \implies x = \frac{1}{2}$$<BR>\n" );
document.write( "There is exactly one critical point at **$x = 0.5$**.\r<BR>\n" );
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document.write( "### 4. Determine the Nature of the Extrema<BR>\n" );
document.write( "We check the second derivative to determine if $x = 0.5$ is a minimum or maximum:<BR>\n" );
document.write( "$$f''(x) = r(r-1)x^{r-2} - r(r-1)(1-x)^{r-2}(-1)$$<BR>\n" );
document.write( "$$f''(x) = r(r-1) \left[ x^{r-2} + (1-x)^{r-2} \right]$$<BR>\n" );
document.write( "Since $r > 1$, $x \in [0, 1]$, and the terms are raised to positive powers, **$f''(x) > 0$** for all $x \in (0, 1)$. This indicates that the function is **convex** (concave up), meaning the critical point at $x = 0.5$ is a **global minimum**.\r<BR>\n" );
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document.write( "### 5. Evaluate the Function at Key Points<BR>\n" );
document.write( "We evaluate $f(x)$ at the critical point and the endpoints of the interval $[0, 1]$:\r<BR>\n" );
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document.write( "* **At the Minimum ($x = 0.5$):**<BR>\n" );
document.write( "    $$f(1/2) = (1/2)^r + (1 - 1/2)^r = (1/2)^r + (1/2)^r$$<BR>\n" );
document.write( "    $$f(1/2) = 2 \cdot \left(\frac{1}{2^r}\right) = \frac{2}{2^r} = \frac{1}{2^{r-1}}$$\r<BR>\n" );
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document.write( "* **At the Endpoints ($x = 0$ and $x = 1$):**<BR>\n" );
document.write( "    $$f(0) = 0^r + (1-0)^r = 1$$<BR>\n" );
document.write( "    $$f(1) = 1^r + (1-1)^r = 1$$\r<BR>\n" );
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document.write( "### 6. Conclusion<BR>\n" );
document.write( "Since the minimum value is $\frac{1}{2^{r-1}}$ and the maximum value at the boundaries is $1$, it follows that for all $x \in [0, 1]$:\r<BR>\n" );
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document.write( "$$\frac{1}{2^{r-1}} \le x^r + (1-x)^r \le 1$$<BR>\n" );
document.write( "$\blacksquare$ </SPAN>\n" );
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