Verify the Mean Value Theorem for the function $f(x) = x^{2} - 4x - 3$ in the interval $[a, b]$,where $a = 1$ and $b = 4$.

(N/A) The given function is $f(x) = x^{2} - 4x - 3$.
Since $f(x)$ is a polynomial function,it is continuous on the closed interval $[1, 4]$ and differentiable on the open interval $(1, 4)$.
The derivative of the function is $f'(x) = 2x - 4$.
Calculate the values at the endpoints:
$f(1) = (1)^{2} - 4(1) - 3 = 1 - 4 - 3 = -6$.
$f(4) = (4)^{2} - 4(4) - 3 = 16 - 16 - 3 = -3$.
According to the Mean Value Theorem,there exists at least one point $c \in (1, 4)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
$\frac{f(4) - f(1)}{4 - 1} = \frac{-3 - (-6)}{3} = \frac{3}{3} = 1$.
Now,set $f'(c) = 1$:
$2c - 4 = 1$.
$2c = 5$.
$c = \frac{5}{2} = 2.5$.
Since $2.5 \in (1, 4)$,the Mean Value Theorem is verified.