91Ó°ÊÓ

If \(f(x)>0\) for every \(x\) in a given interval, the function is entirely above the x-axis within that interval. A function that is entirely above the x-axis would not go below the x-axis (have negative outputs) at any point within that interval.

If \(f'(x)>0\) within an interval, it means that the function's derivative is always positive and the function is always increasing within that interval. The slope of the function is positive, which means its graph would be rising as we move from left to right within the interval.

If \(f''(x)<0\) within an interval, it means the function's second derivative is always negative and the function's rate of increase is decreasing within that interval. That is, the graph of the function would be concave down, indicating that it is becoming less steep as we move from left to right within the interval.

Let's consider the function \(f(x) = -x^2 + 4x + 1\). First, we find the first and second derivatives of the function: \(f'(x) = -2x + 4\) \(f''(x) = -2\) Now, let's analyze the function and its derivatives within the interval \((0, 2)\). 1. \(f(x) = -x^2 + 4x + 1 > 0\) for every \(x\) in the interval \((0, 2)\), which means the function is above the x-axis within the interval. 2. \(f'(x) = -2x + 4 > 0\) for every \(x\) in the interval \((0, 2)\), which means the function is always increasing within the interval. 3. \(f''(x) = -2 < 0\) for every \(x\) in the interval \((0, 2)\), which means the function's rate of increase is decreasing within the interval (concave down). Hence, \(f(x) = -x^2 + 4x + 1\) satisfies all conditions: \(f(x) > 0\), \(f'(x) > 0\), and \(f''(x) < 0\) on the interval \((0, 2)\). Therefore, it is possible for a function to satisfy these conditions on an interval.