Given:
[tex]f(x)=2x^2-3x[/tex]
To determine the derivative using the limit definition, we follow the process shown below:
[tex]\begin{gathered} f(x)=2x^2-3x \\ f^{\prime}(x)=(2x^2)^{\prime}-(3x)^{\prime} \\ \text{Simplify} \\ =2(x^2)^{\prime}-(3x)^{\prime} \\ =2(2)x^{2-1}-3 \\ f^{\prime}(x)=4x-3 \end{gathered}[/tex]
Next, we plug in x=0 into f'(x)=4x-3:
[tex]\begin{gathered} f^{\prime}(0)=4(0)-3 \\ \text{Simplify} \\ f^{\prime}(0)=-3 \end{gathered}[/tex]
Therefore,
[tex]f^{\prime}(0)=-3[/tex]
