Find the rate of change of the function $f( x) =x^{2}-5x$ at:

Solution (a) For $c=2,$ $$f( x) =x^{2}-5x \qquad \hbox{and} \qquad f(2) =2^{2}-5\cdot 2=-6$$

The rate of change of $f$ at $c=2$ is $$\begin{eqnarray*} f^\prime (2) &=& \lim\limits_{x\rightarrow 2}\dfrac{f( x) -f(2) }{x-2}=\lim\limits_{x\rightarrow 2}\dfrac{( x^{2}-5x) -( -6) }{x-2}=\lim\limits_{x\rightarrow 2}\dfrac{ x^{2}-5x+6}{x-2}\\[5pt] &=&\lim\limits_{x\rightarrow 2}\dfrac{( x-2) ( x-3) }{x-2}=\lim\limits_{x\rightarrow 2}\left( x-3\right) =-1 \end{eqnarray*}$$

(b) If $c$ is any real number, then $f( c) =c^{2}-5c$, and the rate of change of $f$ at $c$ is $$\begin{eqnarray*} f^\prime ( c) &=&\lim\limits_{x\rightarrow c}\dfrac{f( x) -f( c) }{x-c}=\lim\limits_{x\rightarrow c}\dfrac{( x^{2}-5x) -( c^{2}-5c) }{x-c}=\lim\limits_{x\rightarrow c} \dfrac{( x^{2}-c^{2}) -5( x-c) }{x-c} \\[5pt] \notag &=&\lim\limits_{x\rightarrow c}\dfrac{( x-c) ( x+c) -5( x-c) }{x-c}=\lim\limits_{x\rightarrow c} \dfrac{( x-c) ( x+c-5) }{x-c}=\lim\limits_{x\rightarrow c} ( x+c-5) \\[3pt] &=&2c-5 \end{eqnarray*}$$