1. **Problem statement:** We have two functions: $$f(x) = \ln(x^2 + 3), \quad x \in \mathbb{R}$$ $$g(x) = \frac{3}{5} + \frac{2}{x + x}, \quad x \in \mathbb{R}, x > -2$$ We need to: (a) State the range of $f$ (b) Find $g^{-1}$ (c) Find $f(g(0))$ (d) Find the exact value of $a$ for which $g(e^{f(e^2)}) = 4a + 3$ --- 2. **(a) Find the range of $f$:** - The function is $f(x) = \ln(x^2 + 3)$. - Since $x^2 \geq 0$ for all real $x$, the minimum value inside the logarithm is when $x=0$, giving $x^2 + 3 = 3$. - The expression inside the logarithm, $x^2 + 3$, ranges from $3$ to $+\infty$. - The natural logarithm function $\ln(y)$ is increasing for $y > 0$, so the minimum value of $f(x)$ is $\ln(3)$ and it increases without bound. **Range of $f$ is:** $$[\ln(3), +\infty)$$ --- 3. **(b) Find $g^{-1}$:** - Given $g(x) = \frac{3}{5} + \frac{2}{x + x} = \frac{3}{5} + \frac{2}{2x} = \frac{3}{5} + \frac{1}{x}$ for $x > -2$. - Let $y = g(x) = \frac{3}{5} + \frac{1}{x}$. - Solve for $x$: $$y - \frac{3}{5} = \frac{1}{x}$$ $$x = \frac{1}{y - \frac{3}{5}} = \frac{1}{y - 0.6}$$ **Therefore,** $$g^{-1}(y) = \frac{1}{y - \frac{3}{5}}$$ --- 4. **(c) Find $f(g(0))$:** - First find $g(0)$: $$g(0) = \frac{3}{5} + \frac{1}{0}$$ - Division by zero is undefined, so $g(0)$ is undefined. **Therefore, $f(g(0))$ is undefined.** --- 5. **(d) Find exact value of $a$ such that $g(e^{f(e^2)}) = 4a + 3$:** - Compute $f(e^2)$: $$f(e^2) = \ln((e^2)^2 + 3) = \ln(e^{4} + 3)$$ - Then compute $e^{f(e^2)}$: $$e^{f(e^2)} = e^{\ln(e^{4} + 3)} = e^{4} + 3$$ - Now compute $g(e^{f(e^2)}) = g(e^{4} + 3)$: $$g(x) = \frac{3}{5} + \frac{1}{x}$$ $$g(e^{4} + 3) = \frac{3}{5} + \frac{1}{e^{4} + 3}$$ - Set equal to $4a + 3$: $$\frac{3}{5} + \frac{1}{e^{4} + 3} = 4a + 3$$ - Solve for $a$: $$4a = \frac{3}{5} + \frac{1}{e^{4} + 3} - 3$$ $$4a = \frac{3}{5} - 3 + \frac{1}{e^{4} + 3} = \frac{3}{5} - \frac{15}{5} + \frac{1}{e^{4} + 3} = -\frac{12}{5} + \frac{1}{e^{4} + 3}$$ $$a = \frac{1}{4} \left(-\frac{12}{5} + \frac{1}{e^{4} + 3}\right) = -\frac{3}{5} + \frac{1}{4(e^{4} + 3)}$$ --- **Final answers:** - (a) Range of $f$: $[\ln(3), +\infty)$ - (b) $g^{-1}(y) = \frac{1}{y - \frac{3}{5}}$ - (c) $f(g(0))$ is undefined - (d) $a = -\frac{3}{5} + \frac{1}{4(e^{4} + 3)}$