We must compute the derivative of the function at x = 2:
[tex]f(x)=4.5x^2-3x+2,[/tex]
using the limit definition.
1) We write explicitly the derivative as a limit:
[tex]f^{\prime}(2)=\lim _{x\rightarrow2}\frac{f(x)-f(2)}{x-2}=\lim _{x\rightarrow2}\frac{(4.5x^2-3x+2)-(4.5\cdot2^2-3\cdot2+2)}{x-2}\text{.}[/tex]
2) We simplify the expression above:
[tex]f^{\prime}(2)=\lim _{x\rightarrow2}\frac{(4.5x^2-3x+2)-14}{x-2}=\lim _{x\rightarrow2}\frac{4.5x^2-3x-12}{x-2}\text{.}[/tex]
If we replace x by 2, we get a 0 in the numerator and denominator. So this is a limit of the type 0/0 where we can apply L'Hopital's rule.
3) We apply L'Hopital's rule, which consists of taking the derivative of numerator and denominator, we get:
[tex]f^{\prime}(2)=\lim _{x\rightarrow2}\frac{4.5\cdot2x^{}-3}{1}\text{.}[/tex]
4) Now, we see that there is no problem to replace x by 2, doing that we get:
[tex]f^{\prime}(2)=4.5\cdot2\cdot2-3=15.[/tex]
Answer: f'(2) = 15
