If (f(x)=2x^2) and (g(x)=x^2+1) then for which (x) will ((f-g)(x)) be zero?

Correct answer: (x=\pm 1)

The direct answer is option A: \(x=\pm1\). To find where a function is zero, set its expression equal to zero. Here \((f-g)(x)=f(x)-g(x)=2x^2-(x^2+1)=x^2-1\). We need \(x^2-1=0\). Rearranging gives \(x^2=1\), so taking both square roots gives \(x=1\) or \(x=-1\), written \(x=\pm1\). Option A is correct. Option B, \(x=0\), is not a zero because \((f-g)(0)=0-1=-1\). Option C, \(x=\pm2\), is not correct because substituting either value gives \(x^2-1=4-1=3\), not zero. Option D is wrong because real solutions do exist, namely 1 and \(-1\). The symbols \(\pm\) are important because both positive and negative numbers have square 1. Memory cue: for zeros, write the function equal to zero before solving.