Problem 63

Question

Determine whether each function is even, odd, or neither. $$g(x)=x^{2}+x$$

Step-by-Step Solution

Verified
Answer
The function \(g(x) = x^{2}+x\) is neither even nor odd.
1Step 1: Calculation of \(g(-x)\)
The function \(g(x)\) is given by \(x^{2}+x\). To find \(g(-x)\), substitute \(-x\) in place of \(x\) in the function: \(g(-x) = (-x)^{2} + (-x) = x^{2} - x\).
2Step 2: Compare \(g(-x)\) with \(g(x)\) and \(-g(x)\)
The function \(g(x)\) is \(x^{2}+x\), and \(g(-x)\) has been found to be \(x^{2} - x\). So, it is clear that \(g(-x) \neq g(x)\) since \(x^{2}+x \neq x^{2} - x\). So, the function \(g(x)\) is not even. Now, let's determine if the function is odd by comparing \(g(-x)\) with \(-g(x)\). The negative of function \(g(x)\) is \(-g(x) = -x^{2} - x\), which is also different from \(g(-x)\), as \(-x^{2} - x \neq x^{2} - x\). Therefore, the function \(g(x)\) is neither even nor odd.