1. **State the problem:** We are given the function $g(x) = x - x^3 + 1$ and want to analyze it. 2. **Write the function:** $$g(x) = x - x^3 + 1$$ 3. **Find the critical points (extrema):** To find extrema, we compute the derivative and set it to zero. $$g'(x) = \frac{d}{dx}(x - x^3 + 1) = 1 - 3x^2$$ 4. **Set derivative equal to zero:** $$1 - 3x^2 = 0$$ 5. **Solve for $x$:** $$3x^2 = 1$$ $$x^2 = \frac{1}{3}$$ $$x = \pm \frac{1}{\sqrt{3}}$$ 6. **Evaluate $g(x)$ at critical points:** $$g\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{\sqrt{3}} - \left(\frac{1}{\sqrt{3}}\right)^3 + 1 = \frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{3}} + 1 = \frac{2}{3\sqrt{3}} + 1$$ $$g\left(-\frac{1}{\sqrt{3}}\right) = -\frac{1}{\sqrt{3}} - \left(-\frac{1}{\sqrt{3}}\right)^3 + 1 = -\frac{1}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + 1 = -\frac{2}{3\sqrt{3}} + 1$$ 7. **Find intercepts:** - **y-intercept:** Set $x=0$: $$g(0) = 0 - 0 + 1 = 1$$ - **x-intercepts:** Solve $g(x) = 0$: $$x - x^3 + 1 = 0$$ $$-x^3 + x + 1 = 0$$ This cubic can be solved numerically or by methods such as the Rational Root Theorem. Testing $x=-1$: $$-(-1)^3 + (-1) + 1 = 1 - 1 + 1 = 1 \neq 0$$ No simple rational roots; approximate numerically if needed. **Summary:** - Critical points at $x = \pm \frac{1}{\sqrt{3}}$ - $y$-intercept at $(0,1)$ - No simple rational $x$-intercepts