Problem 39 - Precalculus Essentials

Question

Determine whether each function is even, odd, or neither. Then determine whether the function's graph is symmetric with respect to the \(y\) -axis, the origin, or neither. $$g(x)=x^{2}+x$$

Step-by-Step Solution

Answer

The function \(g(x)=x^2 + x\) is neither even nor odd. The graph of this function is also not symmetric with respect to either the y-axis or the origin.

Step 1: Substituting \(-x\) into the function

Substitute \(-x\) into each \(x\) in the function \(g(x)\). Doing this, we get \(g(-x) = (-x)^2 + -x = x^2 - x.\)

Step 2: Comparison

Now compare the function \(g(x)=x^2 + x\) with the new function \(g(-x)=x^2 - x\). They are not the same, so we know it's not even. And \(g(-x)\) is not exactly \(g(x)\) with a \(-\) sign at the front, so it is also not odd. Hence, the function is neither even nor odd.

Step 3: Graph Symmetry

We've already found that \(g(x)\) is neither an even nor an odd function. This means the graph of \(g(x)\) is neither symmetric about the y-axis (which is characteristic of even functions) nor about the origin (which is characteristic of odd functions).