26:I[49711,["/_next/static/chunks/0t1_7x_uuxe34.js","/_next/static/chunks/0~ta-e_o9dqvm.js","/_next/static/chunks/115lbazto1qq7.js","/_next/static/chunks/0vq81gpimz.3f.js","/_next/static/chunks/0jal2z51-yxb_.js"],"default"] 27:T478,Both $f(x) = e^x$ and $g(x) = \sqrt{x}$ are continuous for all $x$-values within their domain. The question remains, what are the domains of $f$ and $g$? $f(x) = e^x$ is defined for all real numbers, so its domain is all real numbers. **Therefore, $f$ is continuous for all real numbers.** $g(x) = \sqrt{x}$ is defined for all non-negative numbers, so its domain is all $x$-values such that $x \geq 0$. **Therefore, $g$ is *not* continuous for all real numbers.** **In conclusion, only $f$ is continuous for all real numbers.** --- Licensing information: The question comes from Khan Academy, and is licensed under a creative commons license. The original exercise containing this question can be found here: Continuity and common functions. Note: All Khan Academy content is available for free at www.khanacademy.org. 28:T4a5,Both $h(x) = \cfrac{1}{(x+3)^2}$ and $g(x) = (x-3)^3$ are continuous for all $x$-values within their domain. The question remains, is $x=-3$ within the domains of $h$ and $g$? $h(x) = \cfrac{1}{(x+3)^2}$ is defined for all $x$-values such that $(x+3)^2 \not= 0$. In other words, $h$ is defined for all $x$-values such that $x \not= -3$. **Therefore, $h$ is *not* continuous at $x=-3$.** $g(x) = (x-3)^3$ is defined for all real numbers, so its domain is all real numbers. **Therefore, $g$ is continuous at $x=-3$.** **In conclusion, only $g$ is continuous at $x=-3$.** --- Licensing information: The question comes from Khan Academy, and is licensed under a creative commons license. The original exercise containing this question can be found here: Continuity and common functions. Note: All Khan Academy content is available for free at www.khanacademy.org. 29:T46d,Both $g(x) = 2^x$ and $f(x)= \ln(x)$ are continuous for all $x$-values within their domain. The question remains, what are the domains of $g$ and $f$? $g(x) = 2^x$ is defined for all real numbers, so its domain is all real numbers. **Therefore, $g$ is continuous for all real numbers.** $f(x) = \ln(x)$ is defined for all positive numbers, so its domain is all $x$-values such that $x > 0$. **Therefore, $f$ is *not* continuous for all real numbers.** **In conclusion, only $g$ is continuous for all real numbers.** --- Licensing information: The question comes from Khan Academy, and is licensed under a creative commons license. The original exercise containing this question can be found here: Continuity and common functions. Note: All Khan Academy content is available for free at www.khanacademy.org. 2a:T576,