1. **Problem statement:** Given that $f$ is an increasing function on its domain, determine which of the following functions must also be increasing on their domains: (a) $y = f(x) + 2x$ (b) $y = f(x) - 2x$ (c) $y = \frac{1}{f(x)}$ (d) $y = f(-x)$ 2. **Recall:** A function $g$ is increasing if for any $x_1 < x_2$ in its domain, $g(x_1) < g(x_2)$. 3. **Analyze (a):** Since $f$ is increasing, $f(x_1) < f(x_2)$ for $x_1 < x_2$. Also, $2x$ is increasing because $2x_1 < 2x_2$ for $x_1 < x_2$. Adding two increasing functions results in an increasing function. Therefore, $y = f(x) + 2x$ is increasing. 4. **Analyze (b):** $f(x)$ is increasing, but $-2x$ is decreasing because for $x_1 < x_2$, $-2x_1 > -2x_2$. The sum of an increasing and a decreasing function is not necessarily increasing. For example, if $f(x)$ grows slowly and $-2x$ decreases faster, the sum may decrease. Therefore, $y = f(x) - 2x$ is not necessarily increasing. 5. **Analyze (c):** Consider $y = \frac{1}{f(x)}$. Since $f$ is increasing, $f(x_1) < f(x_2)$ for $x_1 < x_2$. If $f(x)$ is positive and increasing, then $\frac{1}{f(x)}$ is decreasing because the reciprocal of an increasing positive function decreases. If $f(x)$ is negative and increasing (less negative), the reciprocal behavior depends on the sign and domain. Therefore, $y = \frac{1}{f(x)}$ is not necessarily increasing. 6. **Analyze (d):** Consider $y = f(-x)$. For $x_1 < x_2$, $-x_1 > -x_2$. Since $f$ is increasing, $f(-x_1) > f(-x_2)$. Thus, $y = f(-x)$ is decreasing. 7. **Summary:** - (a) must be increasing. - (b), (c), and (d) are not necessarily increasing. **Final answer:** Only (a) $y = f(x) + 2x$ must be increasing on its domain.