Question

let f be the function given by f(x)=\frac{|x^{2}-3|-(x + 0.5)}{(x^{2}-3)(x + 0.5)}. on which of the following open intervals is f continuous? a (-2,-1) b (-1,0) c (0,1) d (1,2)

Explanation:

Step1: Identify points of discontinuity

The function $f(x)=\frac{|x^{2}-3|(x + 0.5)}{(x^{2}-3)(x + 0.5)}$ is discontinuous where the denominator $(x^{2}-3)(x + 0.5)=0$. Solving $x^{2}-3=0$ gives $x=\pm\sqrt{3}\approx\pm1.73$ and solving $x + 0.5=0$ gives $x=- 0.5$.

Step2: Analyze each interval

• For interval A $(-2,-1)$: The values of $x$ in this interval do not make the denominator zero. The function is well - defined and continuous in this interval.
• For interval B $(-1,0)$: $x=-0.5$ is in this interval and makes the denominator zero, so the function is discontinuous in this interval.
• For interval C $(0,1)$: The values of $x$ in this interval do not make the denominator zero, but we need to consider the behavior of the absolute - value function. However, since $x^{2}-3<0$ for $x\in(0,1)$, $|x^{2}-3|=3 - x^{2}$ and the function simplifies. But still, the main concern is the denominator. The function is well - defined in this interval.
• For interval D $(1,2)$: When $x=\sqrt{3}\approx1.73\in(1,2)$, the denominator is zero, so the function is discontinuous in this interval.

Answer:

A. $(-2,-1)$