Question

let f be the function given by f(x) = \frac{|x^{2}-3|\cdot(x + 0.5)}{(x^{2}-3)(x + 0.5)}. on which of the following open intervals is f continuous?

Explanation:

Step1: Identify function's undefined points

The function $f(x)=\frac{|x^{2}-3|(x + 0.5)}{(x^{2}-3)(x + 0.5)}$ is undefined when the denominator $(x^{2}-3)(x + 0.5)=0$.
Set $x^{2}-3=0$, then $x=\pm\sqrt{3}$; set $x + 0.5=0$, then $x=- 0.5$.

Step2: Analyze continuity

The function simplifies to $f(x)=\frac{|x^{2}-3|}{x^{2}-3}$ for $x
eq - 0.5$. The absolute - value function $y = |x^{2}-3|$ and $y=x^{2}-3$ are equal when $x^{2}-3\geq0$ (i.e., $x\leq-\sqrt{3}$ or $x\geq\sqrt{3}$) and $|x^{2}-3|=-(x^{2}-3)$ when $x^{2}-3<0$ (i.e., $-\sqrt{3}<x<\sqrt{3}$). The function $f(x)$ has a removable discontinuity at $x=-0.5$ and non - removable discontinuities at $x = \pm\sqrt{3}$. The function is continuous on the intervals where it is well - defined. So it is continuous on $(-\sqrt{3},-0.5)\cup(-0.5,\sqrt{3})$.

Answer:

The function $f(x)$ is continuous on intervals where it is well - defined. After analyzing the points of discontinuity ($x=\pm\sqrt{3},x = - 0.5$), we need to choose the interval that does not contain these points from the given options (not shown here but the general form of the continuous intervals would be intervals that avoid $x=\pm\sqrt{3},x=-0.5$). For example, if an option is $(-1,1)$ (since $-\sqrt{3}\approx - 1.732$, $\sqrt{3}\approx1.732$ and $-0.5$ is not in $(-1,1)$), it would be a correct choice among the options (assuming such an option exists).