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Section 1-1 : Review : Functions

7. The difference quotient of a function \(f\left( x \right)\) is defined to be,

\[\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]

compute the difference quotient for \(f\left( t \right) = 2{t^2} - 3t + 9\).

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Hint : Don’t get excited about the fact that the function is now \(f\left( t \right)\), the difference quotient still works in the same manner it just has \(t\)’s instead of \(x\)’s now. So, compute \(f\left( {t + h} \right)\), then compute the numerator and finally compute the difference quotient.
Start Solution
\[\begin{align*}f\left( {t + h} \right) & = 2{\left( {t + h} \right)^2} - 3\left( {t + h} \right) + 9 \\ & = 2\left( {{t^2} + 2th + {h^2}} \right) - 3t - 3h + 9\\ & = 2{t^2} + 4th + 2{h^2} - 3t - 3h + 9\end{align*}\] Show Step 2
\[f\left( {t + h} \right) - f\left( t \right) = 2{t^2} + 4th + 2{h^2} - 3t - 3h + 9 - \left( {2{t^2} - 3t + 9} \right) = 4th + 2{h^2} - 3h\] Show Step 3
\[\frac{{f\left( {t + h} \right) - f\left( t \right)}}{h} = \frac{{4th + 2{h^2} - 3h}}{h} = 4t + 2h - 3\]