Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 27E from Chapter 1.1 from Stewart's Calculus, 8th Edition.
Problem 27E
Chapter:
Problem:
Evaluate the difference quotient for the given function. Simplify your answer.
Step-by-Step Solution
Step 1
We are given with following function
\[f\left( x \right) = 4 + 3x - {x^2}\]We have to find the following quotient:\[\dfrac{{f\left( {3 + h} \right) - f\left( 3 \right)}}{h}\]
Step 2: f$(3+h)$
\[\begin{array}{l}f\left( {3 + h} \right) = \left( {4 + 3\left( {3 + h} \right) - {{\left( {3 + h} \right)}^2}} \right)\\ = \left( {4 + 9 + 3h - 9 - 6h - {h^2}} \right)\\ = \left( {4 + 3h - 6h - {h^2}} \right)\\ = \left( {4 - 3h - {h^2}} \right)\end{array}\]Therefore,
\[f\left( {3 + h} \right) = \left( {4 - 3h - {h^2}} \right)\]
Step 3: f$(3)$
\[\begin{array}{l}f\left( 3 \right) = \left( {4 + 3\left( 3 \right) - {{\left( 3 \right)}^2}} \right)\\ = \left( {4 + 9 - 9} \right)\\ = 4\end{array}\]Therefore,
\[f\left( 3 \right) = 4\]
Step 4: Required Quotient f$(3)$
\[\begin{array}{l}\dfrac{{f\left( {3 + h} \right) - f\left( 3 \right)}}{h} = \dfrac{{\left( {4 - 3h - {h^2}} \right) - \left( 4 \right)}}{h}\\ = \dfrac{{ - 3h - {h^2}}}{h}\\ = \dfrac{{h\left( { - 3 - h} \right)}}{h}\\ = - 3 - h\end{array}\]Therefore,
\[\dfrac{{f\left( {3 + h} \right) - f\left( 3 \right)}}{h} = - 3 - h\]
We are given with following function
\[f\left( x \right) = 4 + 3x - {x^2}\]We have to find the following quotient:\[\dfrac{{f\left( {3 + h} \right) - f\left( 3 \right)}}{h}\]
Step 2: f$(3+h)$
\[\begin{array}{l}f\left( {3 + h} \right) = \left( {4 + 3\left( {3 + h} \right) - {{\left( {3 + h} \right)}^2}} \right)\\ = \left( {4 + 9 + 3h - 9 - 6h - {h^2}} \right)\\ = \left( {4 + 3h - 6h - {h^2}} \right)\\ = \left( {4 - 3h - {h^2}} \right)\end{array}\]Therefore,
\[f\left( {3 + h} \right) = \left( {4 - 3h - {h^2}} \right)\]
Step 3: f$(3)$
\[\begin{array}{l}f\left( 3 \right) = \left( {4 + 3\left( 3 \right) - {{\left( 3 \right)}^2}} \right)\\ = \left( {4 + 9 - 9} \right)\\ = 4\end{array}\]Therefore,
\[f\left( 3 \right) = 4\]
Step 4: Required Quotient f$(3)$
\[\begin{array}{l}\dfrac{{f\left( {3 + h} \right) - f\left( 3 \right)}}{h} = \dfrac{{\left( {4 - 3h - {h^2}} \right) - \left( 4 \right)}}{h}\\ = \dfrac{{ - 3h - {h^2}}}{h}\\ = \dfrac{{h\left( { - 3 - h} \right)}}{h}\\ = - 3 - h\end{array}\]Therefore,
\[\dfrac{{f\left( {3 + h} \right) - f\left( 3 \right)}}{h} = - 3 - h\]