91Ó°ÊÓ

First, find the derivative of the function, f'(x), by applying the power rule: $$f'(x)=-8x^{3}+2x$$

The critical points are found by setting the derivative equal to zero and solving for x: $$-8x^{3}+2x=0$$ Factor out x: $$x(-8x^{2}+2)=0$$ Now, solve for x: $$x=0$$ $$-8x^{2}+2=0$$ $$x^{2}=\frac{1}{4}$$ $$x=\pm \frac{1}{2}$$ The critical points are: \(x=-\frac{1}{2}, 0, \frac{1}{2}\).

Create a number line and plot the critical points: $$-\infty \quad -\frac{1}{2} \quad 0 \quad \frac{1}{2} \quad +\infty$$ Choose test points within each interval and evaluate the derivative at those points. The signs of the derivative will indicate whether the function is increasing or decreasing. Interval \((-\infty, -\frac{1}{2})\): choose \(x=-1\). $$f'(-1)=-8(-1)^3+2(-1)=6$$ Since the derivative is positive, the function is increasing on this interval. Interval \((-\frac{1}{2}, 0)\): choose \(x=-\frac{1}{4}\). $$f'(-\frac{1}{4})=-8(-\frac{1}{4})^3+2(-\frac{1}{4})=-\frac{3}{2}$$ Since the derivative is negative, the function is decreasing on this interval. Interval \((0, \frac{1}{2})\): choose \(x=\frac{1}{4}\). $$f'(\frac{1}{4})=-8(\frac{1}{4})^3+2(\frac{1}{4})=-\frac{3}{2}$$ Since the derivative is negative, the function is decreasing on this interval. Interval \((\frac{1}{2}, \infty)\): choose \(x=1\). $$f'(1)=-8(1)^3+2(1)=-6$$ Since the derivative is positive, the function is increasing on this interval. In summary, the function is increasing on the intervals \((-\infty, -\frac{1}{2})\) and \((\frac{1}{2}, \infty)\), and decreasing on the intervals \((-\frac{1}{2}, 0)\) and \((0, \frac{1}{2})\).