Question

Find the least value of ‘a’ so that $f(x) = 2x^2 - ax + 3$ is an increasing function on $[2, 4]$.

Hint:

To find the value of $a$ so that a function is increasing, check the derivative and ensure it is positive over the entire interval. For quadratic functions, focus on the lowest value in the interval.

Answer 1

Given the function: \[f(x) = 2x^2 - ax + 3.\] To find the values of $a$ for which $f(x)$ is increasing on $[2, 4]$, we first compute its derivative: \[f'(x) = \frac{d}{dx} \left( 2x^2 - ax + 3 \right) = 4x - a.\] For $f(x)$ to be increasing on $[2, 4]$, its derivative must be positive for all $x$ in this interval: \[f'(x)>0 \quad \text{for all } x \in [2, 4].\] This implies: \[4x - a>0 \quad \text{for all } x \in [2, 4].\] Considering the smallest value of $x$ in the interval, $x = 2$: \[4(2) - a>0 \quad \Rightarrow \quad 8 - a>0 \quad \Rightarrow \quad a<8.\] Therefore, the least value of $a$ for which $f(x)$ is increasing on $[2, 4]$ is \( a = 8 \).