Given quadratic equation is \(ax^2 + bx + c = 0\) Let \(f'(x) = ax^2 + bx + c\), so \(f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx\). Clearly, \(f(0) = 0\) and \(f(1) = \frac{1}{6}(2a + 3b + 6c) = 0\). Given that \(f(0) = 0 = f(1)\), by Rolle's theorem, \(f'(x)\) has at least one root in \((0, 1)\). The root of \(ax^2 + bx + c = 0\) lies in:
Let \(f(x)\) be a differentiable function on \([0, 8]\) such that \(f(1) = 3\), \(f(2) = 1/2\), \(f(3) = 4\), \(f(4) = -2\), \(f(5) = 6\), \(f(6) = 1/3\), \(f(7) = -1/4\). Then the minimum number of points of interaction of the curve \(y = f'(x)f(x)^2\) and \(y = f'(x)f(x)^2\) is \(k\), then find \(k\).
Let $f:[2,4]\to\mathbb{R}$ be a differentiable function such that $(x\log_e x)f'(x)+(\log_e x)f(x)+f(x)\geq 1,\ x\in[2,4]$ with $f(2)=\dfrac{1}{2}$ and $f(4)=\dfrac{1}{2}$. Consider the following two statements: (A) $f(x)\leq 1$, for all $x\in[2,4]$; (B) $f(x)\geq\dfrac{1}{8}$, for all $x\in[2,4]$. Then,