1. **Problem:** Show that the function $f(x) = x^2 - \cos x$ is an odd function. 2. **Definition:** A function $f$ is odd if $f(-x) = -f(x)$ for all $x$. 3. **Check:** Calculate $f(-x)$: $$f(-x) = (-x)^2 - \cos(-x) = x^2 - \cos x$$ 4. **Compare:** Since $f(-x) = x^2 - \cos x$ and $-f(x) = -(x^2 - \cos x) = -x^2 + \cos x$, we see that $f(-x) \neq -f(x)$. 5. **Conclusion:** Therefore, $f(x) = x^2 - \cos x$ is **not** an odd function. **Note:** The problem asks to show the function is odd, but this function is not odd. --- **Slug:** odd function check **Subject:** algebra **desmos:** {"latex":"","features":{"intercepts":true,"extrema":true}} **q_count:** 9