91Ó°ÊÓ

Using the power rule for differentiation, the derivative of each term in \(f(x)\) can be calculated: $$ f'(x) = -\frac{d}{dx}\left(\frac{x^4}{4}\right) + \frac{d}{dx}\left(x^3\right) - \frac{d}{dx}\left(x^2\right) $$ The derivatives of individual terms are given by: $$ -\frac{d}{dx}\left(\frac{x^4}{4}\right) = -x^3, \quad \frac{d}{dx}\left(x^3\right) = 3x^2, \quad \frac{d}{dx}\left(x^2\right) = 2x $$ So, \(f'(x)\) is given by: $$ f'(x) = -x^3 + 3x^2 - 2x $$

To find the critical points, we need to solve the equation \(f'(x) = 0\): $$ 0 = -x^3 + 3x^2 - 2x $$ The equation has a common factor of \(x\), so we can factor it out: $$ 0 = x(-x^2 + 3x - 2) $$ Now we need to solve the quadratic equation \(-x^2 + 3x - 2 = 0\). Applying the quadratic formula where \(a = -1, b = 3\), and \(c = -2\) yields: $$ x = \frac{-3 \pm \sqrt{(-3)^2 - 4(-1)(-2)}}{2(-1)} $$ Solving for x, we get: $$ x = 1, 2 $$ So, the critical points are \(x = 1\) and \(x = 2\).

Using the critical points \(x=1\) and \(x=2\), we can now divide the number line into three intervals: \((-\infty, 1), (1, 2), (2, +\infty)\). Next, we need to examine the sign of \(f'(x)\) in each of these intervals to determine whether \(f(x)\) is increasing or decreasing. Let's test a number from each interval: 1. For \(x=0\), \(f'(0) = -0^3 + 3(0)^2 - 2(0) = 0\), which lies in \((-\infty, 1)\). In this case, as \(f'(x)=0\) it doesn't give us any information about whether the function slopes upwards or downwards. 2. For \(x=1.5\), \(f'(1.5) = -(1.5)^3 + 3(1.5)^2 - 2(1.5) = 0.375\), which lies in \((1, 2)\). As \(f'(1.5) > 0\), the function is increasing on this interval. 3. For \(x=3\), \(f'(3) = -3^3 + 3(3)^2 - 2(3) = -9\), which lies in \((2, +\infty)\). As \(f'(3) < 0\), the function is decreasing on this interval. Putting it all together, we get: - \(f(x)\) is increasing on the interval \((1, 2)\). - \(f(x)\) is decreasing on the interval \((2, +\infty)\). As visible on the graphs of \(f(x)\) and \(f'(x)\), these intervals match the behavior of the function and its derivative.