1. The problem asks to find $f(-x)$ and $-f(-x)$ for each function $f(x)$, then compare with $f(x)$ to determine if the function is even, odd, or neither. 2. Recall the definitions: - A function is **even** if $f(-x) = f(x)$ for all $x$. - A function is **odd** if $f(-x) = -f(x)$ for all $x$. - Otherwise, the function is **neither**. 3. We analyze each function step-by-step: --- a) $f(x) = x^2 - 4$ - Calculate $f(-x)$: $$f(-x) = (-x)^2 - 4 = x^2 - 4$$ - Calculate $-f(-x)$: $$-f(-x) = -(x^2 - 4) = -x^2 + 4$$ - Compare: $f(-x) = f(x)$ and $-f(-x) \neq f(x)$ - Conclusion: even function. --- b) $f(x) = \sin x + x$ - Calculate $f(-x)$: $$f(-x) = \sin(-x) + (-x) = -\sin x - x$$ - Calculate $-f(-x)$: $$-f(-x) = -(-\sin x - x) = \sin x + x$$ - Compare: $f(-x) \neq f(x)$ but $-f(-x) = f(x)$ - Conclusion: odd function. --- c) $f(x) = \frac{1}{x} - x$ - Calculate $f(-x)$: $$f(-x) = \frac{1}{-x} - (-x) = -\frac{1}{x} + x$$ - Calculate $-f(-x)$: $$-f(-x) = -\left(-\frac{1}{x} + x\right) = \frac{1}{x} - x$$ - Compare: $f(-x) \neq f(x)$ but $-f(-x) = f(x)$ - Conclusion: odd function. --- d) $f(x) = 2x^3 + x$ - Calculate $f(-x)$: $$f(-x) = 2(-x)^3 + (-x) = 2(-x^3) - x = -2x^3 - x$$ - Calculate $-f(-x)$: $$-f(-x) = -(-2x^3 - x) = 2x^3 + x$$ - Compare: $f(-x) \neq f(x)$ but $-f(-x) = f(x)$ - Conclusion: odd function. --- e) $f(x) = 2x^2 - x$ - Calculate $f(-x)$: $$f(-x) = 2(-x)^2 - (-x) = 2x^2 + x$$ - Calculate $-f(-x)$: $$-f(-x) = -(2x^2 + x) = -2x^2 - x$$ - Compare: $f(-x) \neq f(x)$ and $-f(-x) \neq f(x)$ - Conclusion: even or odd neither. --- f) $f(x) = |2x + 3|$ - Calculate $f(-x)$: $$f(-x) = |2(-x) + 3| = |-2x + 3|$$ - Calculate $-f(-x)$: $$-f(-x) = -| -2x + 3|$$ - Compare: $f(-x) \neq f(x)$ and $-f(-x) \neq f(x)$ in general - Conclusion: neither. --- Summary: - a) even - b) odd - c) odd - d) odd - e) neither - f) neither