For \(f(x)=2x^{2}-3x+1\), find the difference quotient \(\dfrac{f(x+h)-f(x)}{h}\), \(h\neq 0\).

Find the limit as \(h\) approaches \(0\) of the difference quotient of \(f( x) =2x^{2}-3x+1\).

Solution (a) To find the difference quotient of \(f\), we begin with \(f( x+h) \). \[ \begin{eqnarray*} f(x+h)&=&2(x+h)^{2}-3(x+h)+1=2( x^{2}+2xh+h^{2}) -3x-3h+1\\[4pt] &=&2x^{2}+4xh+2h^{2}-3x-3h+1 \end{eqnarray*} \]

Now \begin{equation*} f( x+h) -f( x) = ( 2x^{2}+4xh+2h^{2}-3x-3h+1 ) - ( 2x^{2}-3x+1 ) =4xh+2h^{2}-3h \end{equation*}

Then, the difference quotient is \begin{equation*} \dfrac{f(x+h)-f(x)}{h}=\frac{4xh+2h^{2}-3h}{h}=\frac{h(4x+2h-3)}{h}=4x+2h-3, \hbox{ }h\neq 0 \end{equation*}

(b) \(\lim\limits_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h} =\lim\limits_{h\rightarrow 0}( 4x+2h-3) =4x+0-3=4x-3\)