Answer
$$
f(x)=6 x^{2}-4 x
$$
Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&=\lim\limits_{h \to 0}\frac{6 (x+h)^{2}-4 (x+h)-(6 x^{2}-4 x) }{h}\\
&=\lim\limits_{h \to 0}\frac{12 x h+4 h^{2}-4 h}{h}\\
&=\lim\limits_{h \to 0}(12 x+4 h-4 )\\
&=12 x -4.
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=12 (2) -4=20,
$$
$$
f^{\prime}(16)=12 (16) -4=188
$$
and
$$
f^{\prime}(-3)=12 (-3) -4=-40
$$
Work Step by Step
$$
f(x)=6 x^{2}-4 x
$$
Go through the four steps to find $f^{\prime}(x)$
Step 1 Find $f(x+h)$ by replacing $x$ with $x +h$
$$
f(x +h)=6 (x+h)^{2}-4 (x+h)
$$
Step 2
$$
f(x +h) -f(x)=6 (x+h)^{2}-4 (x+h)-(6 x^{2}-4 x)
$$
Step 3
$$
\frac{f(x +h)- f(x)}{h}= \frac{6 (x+h)^{2}-4 (x+h)-(6 x^{2}-4 x) }{h}
$$
Step 4 Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&=\lim\limits_{h \to 0}\frac{6 (x+h)^{2}-4 (x+h)-(6 x^{2}-4 x) }{h}\\
&=\lim\limits_{h \to 0}\frac{12 x h+4 h^{2}-4 h}{h}\\
&=\lim\limits_{h \to 0}(12 x+4 h-4 )\\
&=12 x -4.
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=12 (2) -4=20,
$$
$$
f^{\prime}(16)=12 (16) -4=188
$$
and
$$
f^{\prime}(-3)=12 (-3) -4=-40
$$
