We can find the instantaneous rate of change at [tex]x=2[/tex] by evaluating the limit
[tex]\displaystyle \lim_{x\to2} \frac{f(x) - f(2)}{x - 2}[/tex]
(i.e. the definition of the derivative)
Since
[tex]f(x) = 2x^3 - 3x \implies f(2) = 10[/tex]
we have
[tex]\displaystyle \lim_{x\to2} \frac{f(x) - f(2)}{x - 2} = \lim_{x\to3} \frac{2x^3 - 3x - 10}{x - 2}[/tex]
Note that [tex]2x^3-3x-10=0[/tex] when [tex]x=2[/tex], which means [tex]x-2[/tex] divides the numerator exactly. By polynomial division, we can show
[tex]\dfrac{2x^3 - 3x - 10}{x - 2} = 2x^2 + 4x + 5[/tex]
Then in the limit, we can write
[tex]\displaystyle \lim_{x\to2} \frac{f(x) - f(2)}{x - 2} = \lim_{x\to2} \frac{(x-2) (2x^2 + 4x + 5)}{x - 2}[/tex]
As [tex]x\neq2[/tex], we can cancel the factors of [tex]x-2[/tex].
[tex]\displaystyle \lim_{x\to2} \frac{f(x) - f(2)}{x - 2} = \lim_{x\to2} (2x^2 + 4x + 5)[/tex]
The function in the remaining limit is continuous at [tex]x=2[/tex], so we can directly substitute [tex]x=2[/tex] to get its value,
[tex]\displaystyle \lim_{x\to2} \frac{f(x) - f(2)}{x - 2} = 2\cdot2^2 + 4\cdot2 + 5 = \boxed{21}[/tex]
(D)
