SOLUTION: Let r>1 be a rational number. Prove that for any x->[0,1], 1/2^(r-1) <= x^(r) + (1-x)^(r) <= 1 **Note: I am supposed to use differentiation to find the gradient of an equati

Question 1164420: Let r>1 be a rational number. Prove that for any x->[0,1],
1/2^(r-1) <= x^(r) + (1-x)^(r) <= 1
**Note: I am supposed to use differentiation to find the gradient of an equation in the inequality.
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
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.
### 1. Define the Function
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.
### 2. Find the Gradient (Derivative)
To find the critical points, we take the derivative of $f(x)$ with respect to $x$:
$$f'(x) = rx^{r-1} + r(1-x)^{r-1}(-1)$$
$$f'(x) = r\left[ x^{r-1} - (1-x)^{r-1} \right]$$
### 3. Identify Critical Points
Set the gradient to zero to find the extrema:
$$r\left[ x^{r-1} - (1-x)^{r-1} \right] = 0$$
Since $r > 1$, we divide by $r$:
$$x^{r-1} = (1-x)^{r-1}$$
Taking the $(r-1)$-th root of both sides (which is valid as $x \ge 0$ and $1-x \ge 0$):
$$x = 1 - x \implies 2x = 1 \implies x = \frac{1}{2}$$
There is exactly one critical point at **$x = 0.5$**.
### 4. Determine the Nature of the Extrema
We check the second derivative to determine if $x = 0.5$ is a minimum or maximum:
$$f''(x) = r(r-1)x^{r-2} - r(r-1)(1-x)^{r-2}(-1)$$
$$f''(x) = r(r-1) \left[ x^{r-2} + (1-x)^{r-2} \right]$$
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**.
### 5. Evaluate the Function at Key Points
We evaluate $f(x)$ at the critical point and the endpoints of the interval $[0, 1]$:
* **At the Minimum ($x = 0.5$):**
$$f(1/2) = (1/2)^r + (1 - 1/2)^r = (1/2)^r + (1/2)^r$$
$$f(1/2) = 2 \cdot \left(\frac{1}{2^r}\right) = \frac{2}{2^r} = \frac{1}{2^{r-1}}$$
* **At the Endpoints ($x = 0$ and $x = 1$):**
$$f(0) = 0^r + (1-0)^r = 1$$
$$f(1) = 1^r + (1-1)^r = 1$$
### 6. Conclusion
Since the minimum value is $\frac{1}{2^{r-1}}$ and the maximum value at the boundaries is $1$, it follows that for all $x \in [0, 1]$:
$$\frac{1}{2^{r-1}} \le x^r + (1-x)^r \le 1$$
$\blacksquare$
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