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To determine where the function $f(x)$ is increasing, we need to find its derivative $f'(x)$. $f(x) = 2x^2 - ax + 3$ $f'(x) = 4x - a$
For $f(x)$ to be an increasing function on the interval $[2, 4]$, we must have $f'(x) \ge 0$ for all $x$ in $[2, 4]$. So, $4x - a \ge 0$ for $2 \le x \le 4$.
Since $4x$ is an increasing function, its minimum value on the interval $[2, 4]$ occurs at $x = 2$. Minimum value of $4x$ is $4(2) = 8$.
We need $4x - a \ge 0$, which means $4x \ge a$. Since the minimum value of $4x$ on $[2, 4]$ is 8, we must have $a \le 8$ for $f'(x) \ge 0$ on the entire interval.
We are looking for the least value of $a$ such that $f(x)$ is increasing on $[2, 4]$. Since $a \le 8$, the largest possible value for $a$ is 8. However, the question asks for the LEAST value of 'a' such that the function is increasing on [2,4]. The condition $4x - a \ge 0$ must hold for all $x$ in $[2,4]$. This is equivalent to $a \le 4x$ for all $x$ in $[2,4]$. The smallest value of $4x$ in the interval $[2,4]$ is $4(2) = 8$. Therefore, $a \le 8$. The question is asking for the LEAST value of $a$ such that $f(x)$ is increasing on $[2,4]$. However, there is no least value of $a$ that satisfies the condition $a \le 8$. For example, $a$ could be 7, 6, 5, 0, -1, -100, etc. The question should be asking for the GREATEST value of $a$ such that $f(x)$ is increasing on $[2,4]$. In that case, the answer would be 8. Assuming the question meant to ask for the GREATEST value of 'a', the answer is 8.
Final Answer: 8