Range of Function

Peter, Anne, and Elis were given the task of finding the range of the function: $$ f(x) = x^2 - 2x - 3 $$ with the restricted domain $D(f) = [ 0, 4]$.

First, all three students calculated the function values at the endpoints of the domain and found that $f(0) = -3$ and $f(4) = 5$. Then they proceeded differently:

Peter: Based on the calculated values at the endpoints, he claimed that the range of the function is: $$H(f) = [ -3, 5]$$

Anne: She claimed that the range cannot be determined solely from the function values at the endpoints of the domain, but that it is necessary to calculate the values at other points within the domain, specifically at $x= 1$, $2$, and $3$. Computing theses values, she obtained: $$ f(1) = -4,~f(2) = -3,~f(3) = 0. $$ Since the smallest value she obtained was $-4$, Anne claimed that the range is: $$ H(f) = [ -4, 5] $$

Elis: She rewrote the function $f$ as follows: $$ f(x) = x^2 - 2x - 3 = (x^2 - 2x + 1) - 4 = (x - 1)^2 - 4 $$ She realized that the expression $(x - 1)^2$ attains its minimum value at $x=1$ on the interval $[ 0, 4]$, because this expression is nonnegative for all $x$, and it equals zero at $x = 1$. This expression reaches its maximum at $x=4$ (value $9$) on the interval $[ 0, 4]$. Since $f(1) = -4$ and $f(4) = 5$, Elis concluded that the range is: $$ H(f) = [ -4, 5] $$

Monica: Monica liked how Elis rewrote the function, but her further reasoning was different. She said that the range of the function can be determined from its graph. The graph of this quadratic function is a parabola, and since the coefficient of the quadratic term is positive, the parabola opens upward. Extrema can occur at the vertex of the parabola or at the endpoints of the domain $D(f) = [ 0, 4]$. From the equation: $$ f(x) = (x - 1)^2 - 4 $$ it is clear that the vertex is at $x=1$. So, there are three critical points to check: $$ x = 0,~x = 4,~\mathrm{and}~ x = 1. $$ Monica calculated: $$ f(0) = -3,~f(4) = 5,~\mathrm{and}~f(1) = -4. $$ From these calculations, the minimum value is $-4$ and the maximum value is $5$. Therefore the range is: $$ H(f) = [ -4, 5] $$ Who solved the problem correctly?