1. **Problem Statement:** Find the local maxima and minima of the function $$f(x) = |x^4 - 256|$$ and determine the intervals where the function is increasing or decreasing. 2. **Understanding the function:** The function is the absolute value of $$x^4 - 256$$. The expression inside the absolute value is a quartic polynomial. 3. **Key points:** Set the inside expression to zero to find critical points where the function changes behavior: $$x^4 - 256 = 0 \implies x^4 = 256 \implies x = \pm 4$$ 4. **Rewrite the function piecewise:** - For $$|x| < 4$$, $$x^4 - 256 < 0$$, so $$f(x) = 256 - x^4$$. - For $$|x| \geq 4$$, $$x^4 - 256 \geq 0$$, so $$f(x) = x^4 - 256$$. 5. **Find critical points by differentiating:** - For $$|x| < 4$$: $$f(x) = 256 - x^4$$ $$f'(x) = -4x^3$$ Set $$f'(x) = 0$$: $$-4x^3 = 0 \implies x = 0$$ - For $$|x| > 4$$: $$f(x) = x^4 - 256$$ $$f'(x) = 4x^3$$ Set $$f'(x) = 0$$: $$4x^3 = 0 \implies x = 0$$ (not in this domain) 6. **Evaluate function at critical points and boundaries:** - At $$x=0$$: $$f(0) = |0 - 256| = 256$$ - At $$x=\pm 4$$: $$f(4) = |4^4 - 256| = |256 - 256| = 0$$ $$f(-4) = 0$$ 7. **Determine local maxima and minima:** - At $$x=0$$, $$f(x) = 256$$ is a local maximum because the function decreases on either side (since $$f'(x)$$ changes sign from positive to negative inside $$|x|<4$$). - At $$x=\pm 4$$, $$f(x) = 0$$ are local minima. 8. **Intervals of increase and decrease:** - For $$|x| < 4$$: $$f'(x) = -4x^3$$ - Increasing where $$f'(x) > 0$$: when $$x < 0$$ - Decreasing where $$f'(x) < 0$$: when $$x > 0$$ - For $$|x| > 4$$: $$f'(x) = 4x^3$$ - Increasing where $$f'(x) > 0$$: when $$x > 4$$ - Decreasing where $$f'(x) < 0$$: when $$x < -4$$ - On the intervals $$(-4,4)$$, the function is increasing on $$(-4,0)$$ and decreasing on $$(0,4)$$. 9. **Final answers:** - Local maximum: $$(0, 256)$$ - Local minima: $$( -4, 0 )$$ and $$(4, 0)$$ - Increasing intervals: $$(-4, 0) \cup (4, \infty)$$ - Decreasing intervals: $$(-\infty, -4) \cup (0, 4)$$