If \( f(x)=\frac{\sqrt{x}+2}{3-3 x^{2}} \), for which values of \( x \) is 1.2.2 1.2.3 1.2.4 1. non real. (x) undefined
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To determine where the function \( f(x) = \frac{\sqrt{x} + 2}{3 - 3x^2} \) is undefined, we need to look for values of \( x \) that make the denominator equal to zero. Setting the denominator to zero, we have: \[ 3 - 3x^2 = 0 \] \[ 3x^2 = 3 \] \[ x^2 = 1 \] \[ x = \pm 1 \] So, \( f(x) \) is undefined at \( x = 1 \) and \( x = -1 \). Additionally, since \( \sqrt{x} \) is part of the numerator, \( x \) must also be non-negative for \( \sqrt{x} \) to be real. This means \( x \) must be greater than or equal to 0. Thus, the function \( f(x) \) is undefined at \( x = 1 \) and for any \( x < 0 \), where it does not exist in the real number system. Therefore, the function is non-real for all \( x < 0 \) and it specifically becomes undefined at \( x = 1 \). The final values of \( x \) where \( f(x) \) is undefined are \( x < 0 \) and \( x = 1 \).