91Ó°ÊÓ

First, we'll check if the function is monotonic increasing, which means its first derivative should be positive for all values in its domain. For \(x \geq 1\), \(f(x) = \sqrt{x}\). The first derivative is: \(f'(x) = \frac{1}{2\sqrt{x}}\), which is greater than 0 for all \(x \geq 1\). For \(0 \leq x \leq 1\), \(f(x) = x^3\). The first derivative is: \(f'(x) = 3x^2\), which is greater than or equal to 0 for all \(x \in [0, 1]\). For \(x < 0\), \(f(x) = \frac{x^3}{3}\). The first derivative is: \(f'(x) = x^2\), which is greater than or equal to 0 for all \(x < 0\). So, the derivative is non-negative, but it does not mean that the function is strictly increasing. Thus, statement (a) is false.

Now, we'll check if \(f'(x)\) fails to exist for 3 distinct real values of \(x\). Recall the derivatives of each part: - For \(x \geq 1\): \(f'(x) = \frac{1}{2\sqrt{x}}\) - For \(0 \leq x \leq 1\): \(f'(x) = 3x^2\) - For \(x < 0\): \(f'(x) = x^2\) We know \(f'(x)\) exists for all values of x in their respective ranges, so we only need to check for the points where the function switches between the pieces. At \(x=0\), \(f'(0)=0\). The left-hand and right-hand derivatives are equal, so the function is differentiable at \(x=0\). At \(x=1\), \(f'(1)=\frac{1}{2}\) and \(3(1)^2=3\). The left-hand and right-hand derivatives are different, meaning the function is not differentiable at \(x=1\). Thus, \(f'(x)\) fails to exist for only one distinct real value, so statement (b) is also false.

Statement (c) says that the sign of \(f'(x)\) changes twice. Recall the derivatives of each part one more time: - For \(x \geq 1\): \(f'(x) = \frac{1}{2\sqrt{x}}\) which is positive. - For \(0 \leq x \leq 1\): \(f'(x) = 3x^2\) which is non-negative (positive for all points except 0). - For \(x < 0\): \(f'(x) = x^2\), which is non-negative. As we see, \(f'(x)\) is nonnegative in its entire domain and does not change its sign even once. Therefore, statement (c) is false.

Let's check statement (d), which says that the function attains its extreme values at two positive points \(x_1\) and \(x_2\). Since all derivatives are nonnegative in each region, the function increases in each piece. Therefore, no local maximum exists in any part. The only minimum should be where the function switches between pieces: 1. At \(x=0\), where \(f'(0)=0\), the function has a local minimum (in fact global) since on the left, it is constant, and on the right, it starts to increase. 2. At \(x=1\): this is not an extreme point since there's a discontinuity in the derivative. However, the function is continuous at this point and increases on both sides. Hence, there is no extreme value at this point. Thus, statement (d) is also false. Since all statements are false, none of the given options is true for the function \(f(x)\).