If $$ f(x)=\frac{\sqrt{x+2}}{3-3 x^{2}} $$, for which values of $$ x $$ is 1.2.2 $$ \quad f(x) $$ non real.

Question

UpStudy AI · Accepted Answer

To determine the values of $$ x $$ for which the function
$$
f(x) = \frac{\sqrt{x + 2}}{3 - 3x^2}
$$
is non-real, we need to analyze both the numerator and the denominator:

1. **Numerator ($$ \sqrt{x + 2} $$):**
   - The square root function requires that its argument be non-negative for the function to be real.
   - Therefore, $$ x + 2 \geq 0 $$ implies $$ x \geq -2 $$.
   - If $$ x < -2 $$, the square root becomes non-real (since you would be taking the square root of a negative number).

2. **Denominator ($$ 3 - 3x^2 $$):**
   - The denominator cannot be zero because division by zero is undefined.
   - Setting $$ 3 - 3x^2 = 0 $$ gives:
     $$
     3x^2 = 3 \implies x^2 = 1 \implies x = \pm 1
     $$
   - Thus, $$ x = 1 $$ and $$ x = -1 $$ make the denominator zero, rendering $$ f(x) $$ undefined.

Combining these two conditions:
- **$$ f(x) $$ is non-real if either:**
  - $$ x < -2 $$ (the square root is non-real), **or**
  - $$ x = 1 $$ or $$ x = -1 $$ (denominator is zero).

**Summary:**
The function $$ f(x) $$ is non-real for all real numbers $$ x $$ such that $$ x < -2 $$, $$ x = 1 $$, or $$ x = -1 $$.

**Answer:**
All real numbers below –2 and those equal to –1 or 1. In other words, f (x ) is non real for x < −2 or x = –1 or 1
$$ f(x) $$ is non-real for $$ x < -2 $$, $$ x = -1 $$, or $$ x = 1 $$.

If \( f(x)=\frac{\sqrt{x+2}}{3-3 x^{2}} \), for which values of \( x \) is 1.2.2 \( \quad f(x) \) non real.

Real Tutor Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Pre Calculus Questions