Solve in R :

2. Originally Posted by dhiab
Solve in R :
Hi

The derivative of the function $f(x) = \sqrt{3-x}-\sqrt{x+1}-\frac12$ is negative therefore f is decreasing.

The problem is now to solve $\sqrt{3-x}-\sqrt{x+1}-\frac12 = 0$ which gives $x = 1-\frac{\sqrt{31}}{8}$.

Thus $\sqrt{3-x}-\sqrt{x+1}>\frac12$ over $\left[-1 ; 1-\frac{\sqrt{31}}{8}\right[$

3. Hello, dhiab!

What a bizarre problem . . .

Solve in $\mathbb{R}\!:\;\;\sqrt{3-x} - \sqrt{x+1} \:>\:\tfrac{1}{2}$
We can see that: . $-1 \:\leq \:x\:\leq \:3$

Yet when I solved the inequality algebraically, I got misleading results.

We have: . $2\sqrt{3-x} \:> \:2\sqrt{x+1} + 1$

Square: . $4(3-x) \:>\:4(x+1) + 4\sqrt{x+1} + 1$

Simplify: . $7 - 8x \:>\:4\sqrt{x+1}$

Square: . $49 - 112x + 64x^2 \:>\:16(x+1)$

Simplify: . $64x^2 - 128x + 33 \:>\:0$

This is an upward-opening parabola.
. . When is it above the x-axis?
Find its x-intercepts: . $64x^2 - 128x + 33 \:=\:0$

Quadratic Formula: . $x \;=\;\frac{128 \pm\sqrt{7936}}{128} \;=\;\frac{128\pm16\sqrt{31}}{128} \;=\;\frac{8\pm\sqrt{31}}{8}$

The parabola is positive for: . $\left(x < \frac{8-\sqrt{31}}{8}\right) \;\cup\; \left(x > \frac{8+\sqrt{31}}{8}\right)$

I realize that Squaring can produce extraneous roots,
. . but "half" of this result is inapplicable.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\text{We have: }\;\;\underbrace{\sqrt{3-x}}_{f(x)} \:- \:\underbrace{\sqrt{x+1}}_{g(x)} \:>\:\tfrac{1}{2}$

$f(x)\!:\;\;y \:=\:\sqrt{3-x} \quad\Rightarrow\quad x \:=\:3 - y^2$

This is the upper half of a left-opening parabola with vertex (3,0).

$g(x)\!:\;\;y \:=\:\sqrt{x+1} \quad\Rightarrow\quad x \:=\:y^2 - 1$

This is the upper half of a right-opening parabola with vertex (-1,0).

The question: When is $f(x)$ more than a half-unit above $g(x)$ ?
Code:
                  |
*   .     |                 *
::::*.|         *
::::::|:  *
::::::*       *
:::*  |          *
:*    |            *
- - - * - - + - - - - - - * - - - - -
-1 .    |            . 3
.  |          .
.       .
|   .
. |         .
.         |                 .
|

The solution is the shaded region: . $-1 \leq\:x \:< \:\frac{8-\sqrt{31}}{8}$

Edit: Ha! . . . running-gag neat me to it!
.

4. Blasted arithmetic mistakes!