1. **Problem Statement:** We need to sketch the functions $$f(x) = \frac{x^2 - 9}{x + 3}$$ and $$g(x) = \frac{2x}{x - 1}$$ on the same Cartesian grid. 2. **Identify key features:** - Factorize $$f(x)$$ numerator: $$x^2 - 9 = (x - 3)(x + 3)$$. - Simplify $$f(x)$$ where possible: $$f(x) = \frac{(x - 3)(x + 3)}{x + 3}$$ For $$x \neq -3$$, cancel common factor: $$f(x) = \cancel{\frac{(x - 3)\cancel{(x + 3)}}{\cancel{x + 3}}} = x - 3$$ 3. **Domain and asymptotes:** - $$f(x)$$ has a removable discontinuity (hole) at $$x = -3$$ because of the canceled factor. - $$g(x)$$ has a vertical asymptote at $$x = 1$$ (denominator zero). 4. **Intercepts:** - For $$f(x)$$ (simplified to $$x - 3$$ except at $$x=-3$$): - x-intercept: set $$f(x) = 0$$, $$x - 3 = 0 \Rightarrow x = 3$$ - y-intercept: $$f(0) = 0 - 3 = -3$$ - For $$g(x)$$: - x-intercept: set numerator zero, $$2x = 0 \Rightarrow x = 0$$ - y-intercept: $$g(0) = \frac{0}{-1} = 0$$ 5. **Compare $$f(x) < g(x)$$:** We want to find where: $$\frac{x^2 - 9}{x + 3} < \frac{2x}{x - 1}$$ Multiply both sides by $$(x + 3)(x - 1)$$, noting domain restrictions $$x \neq -3, 1$$: $$ (x^2 - 9)(x - 1) < 2x(x + 3) $$ Expand: $$ (x - 3)(x + 3)(x - 1) < 2x^2 + 6x $$ $$ (x - 3)(x^2 + 2x - 3) < 2x^2 + 6x $$ Expand left side: $$ (x - 3)(x^2 + 2x - 3) = x^3 + 2x^2 - 3x - 3x^2 - 6x + 9 = x^3 - x^2 - 9x + 9 $$ Inequality: $$ x^3 - x^2 - 9x + 9 < 2x^2 + 6x $$ Bring all terms to left: $$ x^3 - x^2 - 9x + 9 - 2x^2 - 6x < 0 $$ $$ x^3 - 3x^2 - 15x + 9 < 0 $$ 6. **Solve cubic inequality:** Find roots of $$x^3 - 3x^2 - 15x + 9 = 0$$ by Rational Root Theorem: Try $$x=3$$: $$3^3 - 3(3)^2 - 15(3) + 9 = 27 - 27 - 45 + 9 = -36 \neq 0$$ Try $$x=1$$: $$1 - 3 - 15 + 9 = -8 \neq 0$$ Try $$x=-1$$: $$-1 - 3 + 15 + 9 = 20 \neq 0$$ Try $$x=9$$: $$729 - 243 - 135 + 9 = 360 \neq 0$$ Try $$x=-3$$: $$-27 - 27 + 45 + 9 = 0$$ So $$x = -3$$ is a root. Divide polynomial by $$x + 3$$: $$x^3 - 3x^2 - 15x + 9 = (x + 3)(x^2 - 6x + 3)$$ Solve quadratic: $$x^2 - 6x + 3 = 0$$ $$x = \frac{6 \pm \sqrt{36 - 12}}{2} = \frac{6 \pm \sqrt{24}}{2} = 3 \pm \sqrt{6}$$ 7. **Sign analysis:** Critical points: $$x = -3, 3 - \sqrt{6}, 3 + \sqrt{6}$$ Test intervals to find where inequality holds: - For $$x < -3$$, test $$x = -4$$: value negative (true) - Between $$-3$$ and $$3 - \sqrt{6}$$, test $$x = 0$$: value positive (false) - Between $$3 - \sqrt{6}$$ and $$3 + \sqrt{6}$$, test $$x = 3$$: value negative (true) - For $$x > 3 + \sqrt{6}$$, test $$x = 5$$: value positive (false) 8. **Domain restrictions:** Exclude $$x = -3$$ and $$x = 1$$ from solution. 9. **Final solution for $$f(x) < g(x)$$:** $$(-\infty, -3) \cup (3 - \sqrt{6}, 3 + \sqrt{6})$$ excluding $$x=1$$ which lies outside these intervals. 10. **Summary for graph:** - Plot $$f(x) = x - 3$$ with a hole at $$x = -3$$. - Plot $$g(x) = \frac{2x}{x - 1}$$ with vertical asymptote at $$x=1$$. - Label intercepts: $$f$$ at (3,0) and (0,-3), $$g$$ at (0,0). - Shade regions on x-axis where $$f(x) < g(x)$$: $$(-\infty, -3)$$ and $$ (3 - \sqrt{6}, 3 + \sqrt{6})$$.