Suppose $f$ is differentiable for all $x$. If $f(1) = -2$ and $f'(x) \geq 2$ for all $x \in [1, 6]$,then:

Rolle's theorem is applicable for the function $f(x) = x^2 - 4$ in which of the following intervals?

If $f(x)=(2x-1)(3x+2)(4x-3)$ is a real-valued function defined on $[\frac{1}{2}, \frac{3}{4}]$, then the value$(s)$ of '$c$' as defined in the statement of Rolle's theorem is/are:

The function $f(x) = (x - 3)^2$ satisfies all the conditions of the Mean Value Theorem in $[3, 4]$. $A$ point on $y = (x - 3)^2$,where the tangent is parallel to the chord joining $(3, 0)$ and $(4, 1)$,is

In the Mean Value Theorem,$f'(c) = \frac{f(b) - f(a)}{b - a}$. If $a = 0$,$b = \frac{1}{2}$,and $f(x) = x(x - 1)(x - 2)$,then the value of $c$ is:

Consider the function $f(x) = 8x^2 - 7x + 5$ on the interval $[-6, 6]$. The value of $c$ that satisfies the conclusion of the Mean Value Theorem is: