1. **State the problem:** We need to find the intervals where the function $$f(x) = x^4 - 6x^3 - 14$$ is increasing. 2. **Recall the rule:** A function is increasing where its derivative $$f'(x)$$ is positive. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(x^4 - 6x^3 - 14) = 4x^3 - 18x^2$$ 4. **Factor the derivative:** $$f'(x) = 2x^2(2x - 9)$$ 5. **Find critical points by setting $$f'(x) = 0$$:** $$2x^2(2x - 9) = 0 \implies x = 0 \text{ or } x = \frac{9}{2} = 4.5$$ 6. **Analyze the sign of $$f'(x)$$ on intervals determined by critical points:** - For $$x < 0$$, choose $$x = -1$$: $$f'(-1) = 2(-1)^2(2(-1) - 9) = 2(1)(-2 - 9) = 2 \times 1 \times (-11) = -22 < 0$$ (decreasing) - For $$0 < x < 4.5$$, choose $$x = 1$$: $$f'(1) = 2(1)^2(2(1) - 9) = 2(1)(2 - 9) = 2 \times 1 \times (-7) = -14 < 0$$ (decreasing) - For $$x > 4.5$$, choose $$x = 5$$: $$f'(5) = 2(5)^2(2(5) - 9) = 2(25)(10 - 9) = 2 \times 25 \times 1 = 50 > 0$$ (increasing) 7. **Conclusion:** The function $$f(x)$$ is increasing on the interval $$\boxed{(4.5, \infty)}$$. --- **Next, find relative maxima of $$f$$ on $$(-9,9)$$ using the graph of $$f'$$:** - Relative maxima occur where $$f'$$ changes from positive to negative. - From the graph description, $$f'$$ crosses zero near $$x = -3$$ going from positive to negative (peak), and near $$x = 5$$ going from positive to negative (peak). Therefore, relative maxima of $$f$$ occur near $$x = -3$$ and $$x = 5$$ on $$(-9,9)$$. --- **Summary:** - $$f(x)$$ is increasing on $$\boxed{(4.5, \infty)}$$. - Relative maxima of $$f$$ on $$(-9,9)$$ are near $$x = -3$$ and $$x = 5$$.