Evaluating a Function

For the function \(f\) defined by \(f(x) =2x^{2}-3x\), find:

1. \(f( 5) \)
2. \(f( x+h)\)
3. \(f( x+h) -f( x) \)
4. \(\dfrac{f( x+h) -f( x) }{h}, h\neq 0\)

Solution (a) \(f( 5) =2 ( 5)^{2}-3( 5) =50-15=35\)

(b) The function \(f( x) =2x^{2}-3x\) gives us a rule to follow. To find \(f( x+h) \), expand \(( x+h) ^{2} \), multiply the result by \(2\), and then subtract the product of \(3\) and \(( x+h)\). \[ \begin{eqnarray*} &&f( x+h) = 2( x+h) ^{2}-3( x+h) =2( x^{2}+2hx+h^{2}) -3x-3h\\ &&\,\underset{\color{#0066A7}{\hbox{In}\ f( x)\ \hbox{replace}\ x\ \hbox{by}\ x+h}}{\color{#0066A7}{{\uparrow}}}\\ &&\hspace{5.5pc} =2x^{2}+4hx+2h^{2}-3x-3h \end{eqnarray*} \]

(c) \(f( x+h) -f( x) =[ 2x^{2}+4hx+2h^{2}-3x-3h] -[ 2x^{2}-3x] =\) \(4hx+2h^{2}-3h\)

(d) \[ \begin{eqnarray*} &&\dfrac{f( x+h) -f( x) }{h}= \dfrac{4hx+2h^{2}-3h}{h}=\dfrac{h[ 4x+2h-3] }{h} \underset{\underset{\color{#0066A7}{h \neq 0;\ \hbox{divide out}\ h}}{\color{#0066A7}{\uparrow}}}{=} 4x+2h-3\\ \end{eqnarray*} \]