Answer:
Recall that, if the limit exists:
[tex]f^{\prime}(x)=\lim _{h\rightarrow0}\frac{f(x+h)-f(x)}{h}.[/tex]
Now, we compute f(x+h):
[tex]\begin{gathered} f(x+h)=4.5(x+h)^2-3(x+h)+2 \\ =4.5(x^2+2xh+h^2)-3x-3h+2 \\ =4.5x^2+9xh+4.5h^2-3x-3h+2 \\ =4.5x^2-3x+2+9xh+4.5h^2-3h \\ =f(x)+9xh+4.5h^2-3h\text{.} \end{gathered}[/tex]
Therefore:
[tex]\begin{gathered} \lim _{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=\lim _{h\rightarrow0}\frac{f(x)+9xh+4.5h^2-3h-f(x)}{h} \\ =\lim _{h\rightarrow0}\frac{9xh+4.5h^2-3h}{h}=\lim _{h\rightarrow0}(9x+4.5h^{}-3) \\ =9x-3. \end{gathered}[/tex]
Therefore:
[tex]f^{\prime}(x)=9x-3.[/tex]
