Domain of a function

[grapz]

y = 1/2 ln ( 3 - x^2)

so 3 - x ^2 > 0
x ^ 2 < 3

now the answer says that the |x| < 3^1/2 --> -3^1/2 < x < 3 ^ 1/2 --> 0 <_ x < 3 ^ 1/2.

so the final answer is [0 , 3 ^1/2)


The question is, why is it |X| < 3^1/2 and not x < 3^1/2. And why is it 0 <_ x < 3 ^1/2, instead of -3 ^1/2 < x < 3 ^ 1/2

thx

 

[skeeter]

$\displaystyle 3-x^2 > 0$

3 - x² equals 0 at $\displaystyle x = \pm \sqrt{3}$.

these two values of x break up the break up the real number line into three sections where x is not 0 ... $\displaystyle x < -\sqrt{3}$, $\displaystyle \, -\sqrt{3} < x < \sqrt{3}$, and $\displaystyle x > \sqrt{3}$.

pick a value in each section, and "test" it in the original inequality ...

for $\displaystyle x < -\sqrt{3}$ ... let x = -2 ... 3 - (-2)² = -1, which is not greater than 0. so ... every value in this section makes the original inequality false.

for $\displaystyle \, -\sqrt{3} < x < \sqrt{3}$ ... let x = 0 ... 3 - (0)² = 3, which is greater than 0. so ... every x-value in this interval makes the original inequality true.

the test for the last interval $\displaystyle x > \sqrt{3}$ also proves to be false.

therefore, the domain of the log function is $\displaystyle \, -\sqrt{3} < x < \sqrt{3}$, which also can be written as $\displaystyle |x| < \sqrt{3}$.

fyi ... one can also "visualize" where 3 - x² is greater than 0 by looking at the graph of the parabola $\displaystyle y = 3 - x^2$, an inverted parabola with vertex at (0,3) and y-intercepts at $\displaystyle (\sqrt{3}, 0)$ and $\displaystyle (-\sqrt{3}, 0)$ ... where does the graph have values of x where the y-values are positive?