1. **State the problem:** We are given two functions: $$h(x) = \sqrt{\frac{e^{3x}}{x^2 + 1}}$$ and $$k(x) = \sqrt[3]{4^{5x} \ln(x^2)}$$ We want to understand these functions, their components, and how they behave. 2. **Recall relevant formulas and rules:** - The square root function is defined as $$\sqrt{a} = a^{\frac{1}{2}}$$. - The cube root function is defined as $$\sqrt[3]{a} = a^{\frac{1}{3}}$$. - Exponent rules: $$a^{m} \cdot a^{n} = a^{m+n}$$ and $$\frac{a^{m}}{a^{n}} = a^{m-n}$$. - The natural logarithm $$\ln(x)$$ is defined for $$x > 0$$. 3. **Analyze and simplify $$h(x)$$:** $$h(x) = \sqrt{\frac{e^{3x}}{x^2 + 1}} = \left(\frac{e^{3x}}{x^2 + 1}\right)^{\frac{1}{2}} = \frac{(e^{3x})^{\frac{1}{2}}}{(x^2 + 1)^{\frac{1}{2}}} = \frac{e^{\frac{3x}{2}}}{\sqrt{x^2 + 1}}$$ 4. **Analyze and simplify $$k(x)$$:** $$k(x) = \sqrt[3]{4^{5x} \ln(x^2)} = \left(4^{5x} \ln(x^2)\right)^{\frac{1}{3}} = (4^{5x})^{\frac{1}{3}} (\ln(x^2))^{\frac{1}{3}} = 4^{\frac{5x}{3}} (\ln(x^2))^{\frac{1}{3}}$$ 5. **Rewrite base 4 as powers of 2 for clarity:** Since $$4 = 2^2$$, $$4^{\frac{5x}{3}} = (2^2)^{\frac{5x}{3}} = 2^{\frac{10x}{3}}$$ So, $$k(x) = 2^{\frac{10x}{3}} (\ln(x^2))^{\frac{1}{3}}$$ 6. **Domain considerations:** - For $$h(x)$$, the denominator $$\sqrt{x^2 + 1}$$ is always positive, so $$h(x)$$ is defined for all real $$x$$. - For $$k(x)$$, $$\ln(x^2)$$ requires $$x^2 > 0$$, so $$x \neq 0$$. 7. **Summary:** - $$h(x) = \frac{e^{\frac{3x}{2}}}{\sqrt{x^2 + 1}}$$ - $$k(x) = 2^{\frac{10x}{3}} (\ln(x^2))^{\frac{1}{3}}$$ These forms show the exponential growth and logarithmic components clearly.