91Ó°ÊÓ

To find the first derivative of the function, we will use the power rule. Let's find the derivative of f(x) with respect to x: $$f'(x)=-8x^3+2x$$

Critical points are the points where the first derivative is equal to zero or it's undefined (the latter doesn't apply in this case). Let's find the critical points by setting f'(x) equal to zero and solving for x: $$-8x^3+2x = 0$$ Factor out a common factor of 2x: $$2x(-4x^2+1) = 0$$ Now, we have two factors, and set them equal to zero: $$2x=0 \Longrightarrow x=0$$ $$-4x^2 + 1 = 0 \Longrightarrow x^2 = \frac{1}{4} \Longrightarrow x = \pm\frac{1}{2}$$ So, we have three critical points: \(x = -\frac{1}{2}, x = 0, x = \frac{1}{2}\).

To determine the intervals of increase and decrease, we'll test the sign of the first derivative within each interval between the critical points: 1. Interval \((-\infty, -\frac{1}{2})\): Choose a test point, for example \(x=-1\). Plug into the first derivative: $$f'(-1) = -8(-1)^3 + 2(-1) = -10 < 0$$ Since f'(-1) < 0, the function is decreasing in this interval. 2. Interval \((-\frac{1}{2}, 0)\): Choose a test point, for example \(x=-\frac{1}{4}\). Plug into the first derivative: $$f'\left(-\frac{1}{4}\right) = -8\left(-\frac{1}{4}\right)^3+2\left(-\frac{1}{4}\right) = 1 > 0$$ Since f'(-1/4) > 0, the function is increasing in this interval. 3. Interval \((0, \frac{1}{2})\): Choose a test point, for example \(x=\frac{1}{4}\). Plug into the first derivative: $$f'\left(\frac{1}{4}\right) = -8\left(\frac{1}{4}\right)^3+2\left(\frac{1}{4}\right)= 1 > 0$$ Since f'(1/4) > 0, the function is increasing in this interval. 4. Interval\((\frac{1}{2},\infty)\): Choose a test point, for example \(x=1\). Plug into the first derivative: $$f'(1) = -8(1)^3 + 2(1) = -6 < 0$$ Since f'(1) < 0, the function is decreasing in this interval. In conclusion, the function is increasing on the intervals \((-\frac{1}{2},0)\) and \((0,\frac{1}{2})\) and decreasing on the intervals \((-\infty,-\frac{1}{2})\) and \((\frac{1}{2},\infty)\).