1. State the problem. We are given six functions and must compute $f(-x)$ and $-f(-x)$ for each, then compare with $f(x)$ to decide whether the function is even, odd, or neither. 2. a) For $f(x)=x^2-4$ compute $f(-x)$. $$f(-x)=(-x)^2-4=x^2-4$$ Then compute $-f(-x)$. $$-f(-x)=-(x^2-4)=-x^2+4$$ Compare $f(-x)$ with $f(x)$ and $-f(-x)$. We have $f(-x)=x^2-4=f(x)$, so $f$ is even. 3. b) For $f(x)=\sin x + x$ compute $f(-x)$. $$f(-x)=\sin(-x)+(-x)=-\sin x - x$$ This simplifies to $f(-x)=-(\sin x + x)=-f(x)$. Therefore $f$ is odd. 4. c) For $f(x)=\frac{1}{x}-x$ note the domain excludes $x=0$. Compute $f(-x)$. $$f(-x)=\frac{1}{-x}-(-x)=-\frac{1}{x}+x$$ This equals $-(\frac{1}{x}-x)$, so $f(-x)=-f(x)$. Therefore $f$ is odd on its domain $x\neq 0$. 5. d) For $f(x)=2x^3+x$ compute $f(-x)$. $$f(-x)=2(-x)^3+(-x)=-2x^3-x$$ This is $-(2x^3+x)=-f(x)$, so $f$ is odd. 6. e) For $f(x)=2x^2-x$ compute $f(-x)$. $$f(-x)=2(-x)^2-(-x)=2x^2+x$$ Compute $-f(-x)$ for comparison. $$-f(-x)=-(2x^2+x)=-2x^2-x$$ Compare with $f(x)=2x^2-x$. Neither $f(-x)=f(x)$ nor $f(-x)=-f(x)$ holds for all $x$, so $f$ is neither even nor odd. 7. f) For $f(x)=|2x+3|$ compute $f(-x)$. $$f(-x)=|2(-x)+3|=|-2x+3|=|3-2x|$$ Compare $|3-2x|$ with $|2x+3|$ and with $-|3-2x|$. Absolute values are nonnegative, so $-f(-x)$ cannot equal $f(x)$ unless both are zero for all $x$, which is false. A simple counterexample is $x=1$: $f(1)=|5|=5$ and $f(-1)=|1|=1$, so $f(-1)\neq f(1)$ and $f(-1)\neq -f(1)$. Therefore $f$ is neither even nor odd. Final answers. a) even. b) odd. c) odd (for $x\neq 0$). d) odd. e) neither. f) neither.