Question

Find the domain of the function. $$ f(x)=\frac{\sqrt{x+2}+\sqrt{2-x}}{x^{3}-x} $$

Short answer

The domain of the function \(f(x)=\frac{\sqrt{x+2}+\sqrt{2-x}}{x^{3}-x}\) is \(Domain(f(x)) = \left[-2, -1\right) \cup \left(-1, 0\right) \cup \left(0, 1\right) \cup \left(1, 2\right]\).

Step-by-step solution

Identifying problematic square roots

First, let's identify the square roots that may cause issues in the function. The square root of a negative number is not defined for real numbers. Thus, we need to consider when the expressions under the square root symbols are greater than or equal to zero: \(x+2 \geq 0\) and \(2-x \geq 0\)

Solve the inequalities for x

Solving the inequalities from step 1, to determine the x values which make both square roots valid: For \(x + 2 \geq 0\): Subtract 2 from both sides: \(x \geq -2\) For \(2 - x \geq 0\): Add x to both sides and subtract 2: \(x \leq 2\) By combining these inequalities, we get: \(-2 \leq x \leq 2\)

Identifying potential division by zero

Next, we need to consider when the denominator is equal to zero, as division by zero is undefined. To do this, set the denominator equal to zero and solve for x: \(x^3 - x = 0\)

Factor and solve for x

In order to find the x values that result in division by zero in the denominator, we can factor the equation from step 3. Factoring, we get: \(x(x^2 - 1) = 0\) Now we have a product equal to zero, so one of the factors must be zero: \(x = 0\) or \(x^2 - 1 = 0\) For \(x^2 - 1 = 0\), we can add 1 to both sides and take the square root: \(x = \pm 1\) So, the denominator is zero when \(x = -1, 0,\) or \(1\).

Removing undefined x values from the domain

Since the denominator is zero when \(x = -1, 0,\) or \(1\), these x values are not in the domain of the function. So, we must remove them from the range \(-2 \leq x \leq 2\) that we found earlier: \(-2 \leq x < -1\) or \(-1 < x < 0\) or \(0 < x < 1\) or \(1 < x \leq 2\)

Write the domain of the function

Finally, we can present the domain of the function as a union of the intervals from Step 5: \( Domain(f(x)) = \left[-2, -1\right) \cup \left(-1, 0\right) \cup \left(0, 1\right) \cup \left(1, 2\right] \)