Question

Number of integer in the domain of the function

$f(x)=\frac{\sqrt{9-x^2}}{\log(x+4)}$, is

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (C)

6

Answer 2

The domain of the function $f(x)=\frac{\sqrt{9-x^2}}{\log(x+4)}$ is determined by the following conditions:

1. The expression under the square root must be non-negative: $9-x^2 \ge 0$
   $x^2 \le 9$
   $-3 \le x \le 3$

2. The argument of the logarithm must be positive: $x+4 > 0$
   $x > -4$

3. The denominator cannot be zero: $\log(x+4) \ne 0$
   This implies $x+4 \ne 1$
   $x \ne -3$

To find the domain of $f(x)$, we must satisfy all three conditions simultaneously. We find the intersection of the intervals obtained from these conditions.

From condition 1: $x \in [-3, 3]$.
From condition 2: $x \in (-4, \infty)$.
From condition 3: $x \ne -3$.

The intersection of $[-3, 3]$ and $(-4, \infty)$ is $[-3, 3]$.
Now, we apply the condition $x \ne -3$ to the interval $[-3, 3]$. This excludes the point $x=-3$ from the interval.
The resulting domain is $(-3, 3]$.

We are asked to find the number of integers in the domain $(-3, 3]$.
The integers in this interval are the integers $x$ such that $-3 < x \le 3$.
The integers satisfying this condition are $-2, -1, 0, 1, 2, 3$.

Counting these integers, we find there are 6 integers.