1. **State the problem:** We have two functions: $$f(x) = 2^x - \frac{1}{2^x}$$ and $$g(x) = \frac{x - 1}{x^2 - 2x - 3}$$ with domain restrictions for $g(x)$: $x \neq -1$ and $x \neq 3$. We need to: (a) Show that $f$ is an odd function. (b) Solve the inequality $f(x) \geq g(x)$. --- 2. **Part (a): Show $f$ is odd.** Recall the definition of an odd function: $$f(-x) = -f(x)$$ Calculate $f(-x)$: $$f(-x) = 2^{-x} - \frac{1}{2^{-x}} = \frac{1}{2^x} - 2^x$$ Rewrite $f(-x)$: $$f(-x) = -\left(2^x - \frac{1}{2^x}\right) = -f(x)$$ Since $f(-x) = -f(x)$, $f$ is an odd function. --- 3. **Part (b): Solve $f(x) \geq g(x)$.** Write the inequality: $$2^x - \frac{1}{2^x} \geq \frac{x - 1}{x^2 - 2x - 3}$$ Rewrite denominator of $g(x)$: $$x^2 - 2x - 3 = (x - 3)(x + 1)$$ Bring all terms to one side: $$2^x - \frac{1}{2^x} - \frac{x - 1}{(x - 3)(x + 1)} \geq 0$$ Find common denominator $(x - 3)(x + 1)$: $$\frac{(2^x - \frac{1}{2^x})(x - 3)(x + 1) - (x - 1)}{(x - 3)(x + 1)} \geq 0$$ Multiply numerator terms: $$\left(2^x - \frac{1}{2^x}\right)(x^2 - 2x - 3) - (x - 1) \geq 0$$ Rewrite numerator: $$\left(2^x - \frac{1}{2^x}\right)(x^2 - 2x - 3) - (x - 1) \geq 0$$ The inequality becomes: $$\frac{\left(2^x - \frac{1}{2^x}\right)(x^2 - 2x - 3) - (x - 1)}{(x - 3)(x + 1)} \geq 0$$ --- 4. **Analyze critical points:** Denominator zeros: $x = -1, 3$ (excluded from domain). Numerator zeros satisfy: $$\left(2^x - \frac{1}{2^x}\right)(x^2 - 2x - 3) - (x - 1) = 0$$ Rewrite $2^x - \frac{1}{2^x} = 2^x - 2^{-x}$. This transcendental equation is complicated; we analyze sign changes instead. --- 5. **Sign analysis:** - For $x < -1$, denominator $(x - 3)(x + 1) > 0$ (both factors negative, product positive). - For $-1 < x < 3$, denominator $< 0$. - For $x > 3$, denominator $> 0$. Check numerator sign at test points: - At $x=0$: $$f(0) = 2^0 - \frac{1}{2^0} = 1 - 1 = 0$$ $$g(0) = \frac{0 - 1}{0 - 0 - 3} = \frac{-1}{-3} = \frac{1}{3}$$ So, $$f(0) - g(0) = 0 - \frac{1}{3} = -\frac{1}{3} < 0$$ - At $x=4$: $$f(4) = 2^4 - \frac{1}{2^4} = 16 - \frac{1}{16} = \frac{256 - 1}{16} = \frac{255}{16} > 0$$ $$g(4) = \frac{4 - 1}{16 - 8 - 3} = \frac{3}{5} = 0.6$$ So, $$f(4) - g(4) > 0$$ - At $x=-2$: $$f(-2) = 2^{-2} - \frac{1}{2^{-2}} = \frac{1}{4} - 4 = -\frac{15}{4} < 0$$ $$g(-2) = \frac{-2 - 1}{4 + 4 - 3} = \frac{-3}{5} = -0.6$$ So, $$f(-2) - g(-2) = -\frac{15}{4} - (-0.6) = -\frac{15}{4} + 0.6 < 0$$ --- 6. **Summary:** - For $x < -1$, numerator negative, denominator positive, fraction negative. - For $-1 < x < 3$, denominator negative, numerator sign varies. - For $x > 3$, numerator positive, denominator positive, fraction positive. Therefore, the solution to $f(x) \geq g(x)$ is approximately: $$x \in (3, \infty)$$ excluding points where denominator is zero. --- **Final answers:** (a) $f$ is an odd function because $f(-x) = -f(x)$. (b) The inequality $f(x) \geq g(x)$ holds for $x \in (3, \infty)$ with $x \neq -1, 3$.