If $f(x) = x^2 - 2x + 4$ and $\frac{f(5) - f(1)}{5 - 1} = f'(c)$,then the value of $c$ will be

For all $x \in [0, 2024]$,assume that $f(x)$ is differentiable,$f(0) = -2$,and $f^{\prime}(x) \geq 5$. Then the least possible value of $f(2024)$ is:

Let $y = f(x)$ and $y = g(x)$ be two differentiable functions in $[0, 2]$ such that $f(0) = 3$,$f(2) = 5$,$g(0) = 1$,and $g(2) = 2$. If there exists at least one $c \in (0, 2)$ such that $f'(c) = k g'(c)$,then $k$ must be:

If from the Mean Value Theorem,$f'({x_1}) = \frac{f(b) - f(a)}{b - a}$,then

$A$ value of $c$ according to the Lagrange's mean value theorem for $f(x)=(x-1)(x-2)(x-3)$ in $[0,4]$ is

Let $f(x)$ be continuous on $[0,6]$ and differentiable on $(0,6)$. Let $f(0)=12$ and $f(6)=-4$. If $g(x)=\frac{f(x)}{x+1}$, then for some Lagrange's constant $c \in(0,6)$, $g^{\prime}(c)=$