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Chapter 4: Problem 2

Complete the following table of function values and hence sketch an accurate graph of \(f(x)=x^{3}\). $$ \begin{array}{lccccccc} x & -1.50 & -1.25 & -1.00 & -0.75 & -0.50 & -0.25 & 0.00 \\ \hline f(x) & & -1.95 & & & -0.13 & & \\ x & 0.25 & 0.50 & 0.75 & 1.00 & 1.25 & 1.50 & \\ \hline f(x) & & 0.13 & & & 1.95 & & \end{array} $$ Draw the tangents to the graph at \(x=-1,0\) and 1 . Hence estimate the values of \(f^{\prime}(-1), f^{\prime}(0)\) and \(f^{\prime}(1)\).

Short Answer

Expert verified
The completed table: \[ \begin{array}{lccccccc} x & -1.50 & -1.00 & -0.75 & -0.25 & 0.00 & 0.25 & 1.00 & 1.50 \ f(x) & -3.375 & -1.00 & -0.421875 & -0.015625 & 0.00 & 0.015625 & 1.00 & 3.375 \end{array} \]Estimated derivatives: \[ f'(-1) \approx 3 \], \[ f'(0) = 0 \], \[ f'(1) \approx 3 \].

Step by step solution

01

Understanding the Function

The function given is a cubic function: \[ f(x) = x^3 \]. A cubic function has an S-shaped curve.
02

Complete the Table

Calculate the values of the function for the given x values: \[ f(-1.50) = (-1.50)^3 = -3.375 \]\[ f(-1.00) = (-1.00)^3 = -1.00 \]\[ f(-0.75) = (-0.75)^3 = -0.421875 \]\[ f(-0.25) = (-0.25)^3 = -0.015625 \]\[ f(0.00) = (0.00)^3 = 0.00 \]\[ f(0.25) = (0.25)^3 = 0.015625 \]\[ f(0.75) = (0.75)^3 = 0.421875 \]\[ f(1.00) = (1.00)^3 = 1.00 \]\[ f(1.50) = (1.50)^3 = 3.375 \].Update the table with these values.
03

Updated Table

The table is now complete as shown below:\[\begin{array}{lccccccc} x & -1.50 & -1.25 & -1.00 & -0.75 & -0.50 & -0.25 & 0.00 \f(x) & -3.375 & -1.95 & -1.00 & -0.421875 & -0.13 & -0.015625 & 0.00 \ x & 0.25 & 0.50 & 0.75 & 1.00 & 1.25 & 1.50 & \f(x) & 0.015625 & 0.13 & 0.421875 & 1.00 & 1.95 & 3.375 & .\end{array}\]
04

Sketch the Graph

Plot the points \[ (-1.50, -3.375), (-1.25, -1.95), (-1.00, -1.00), (-0.75, -0.421875), (-0.50, -0.13), (-0.25, -0.015625), (0.00, 0.00), (0.25, 0.015625), (0.50, 0.13), (0.75, 0.421875), (1.00, 1.00), (1.25, 1.95), (1.50, 3.375) \] on a graph and connect these points smoothly to form the curve of \[ f(x) = x^3 \].
05

Draw Tangents

Draw tangents to the graph at points \(x = -1, 0, 1\). A tangent line is a line that touches the curve at a point without crossing it.
06

Estimate Derivatives

The slope of the tangent line at any point on the graph of a function gives the value of the derivative at that point. Estimate the slopes as follows:For \(x = -1\): The tangent has a slope of approximately 3. \(f'(-1) \approx 3\).For \(x = 0\): The tangent line is horizontal, so the slope is 0. \(f'(0) = 0\).For \(x = 1\): The tangent has a slope of approximately 3. \(f'(1) \approx 3\).
07

Final Answer

Compile all the findings into a coherent answer.\[ f(-1.5) = -3.375 \]\[ f(-1.00) = -1.00 \]\[ f(-0.75) = -0.421875 \]\[ f(-0.25) = -0.015625 \]\[ f(0.00) = 0.00 \]\[ f(0.25) = 0.015625 \]\[ f(0.75) = 0.421875 \]\[ f(1.00) = 1.00 \]\[ f(1.50) = 3.375 \].Estimated derivatives are: \[ f'(-1) \approx 3 \], \[ f'(0) = 0 \], \[ f'(1) \approx 3 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Table Completion
To fully understand a function like the cubic function given, we need to complete its function table first. The function provided is \( f(x) = x^3 \). Completing the table involves calculating the function鈥檚 value for each x given. Here鈥檚 how you can do it:
For each \( x \) value in the table, substitute it into the function to find \( f(x) \).
For instance:
\( f(-1.50) = (-1.50)^3 = -3.375 \)
\( f(1.50) = (1.50)^3 = 3.375 \)
By following this process, you fill in the corresponding \( f(x) \) values for every \( x \) given. It is crucial that every calculation is checked to ensure accuracy.
The completed table allows us to move to the graphing stage more efficiently.
Graph Sketching
Once the function table is complete, the next step is to sketch the graph. This visual representation helps us see the behavior of the function. Here鈥檚 a simple guide:
Start by plotting the points from the table onto a coordinate system. Each \( x \) value from the table corresponds to a point on the graph.
For example, plot the points \( (-1.50, -3.375) \), \( (-1.00, -1.00) \), and so on.
Connect these points smoothly. Polynomial functions like cubic ones have continuous, smooth curves.
As you connect the points, you'll notice the characteristic shape of the cubic function, which is an S-shaped curve. This visualization provides a solid base to understand how the function behaves.
Tangent Lines
The final part of the exercise involves drawing tangents to the graph and estimating derivatives. A tangent line touches the graph at exactly one point without crossing it.
To draw a tangent:
  • Identify the points at \( x = -1, 0, \text{and } 1 \) on your graph.
  • Draw lines that touch the curve at these specific points.
  • Ensure the lines extend in both directions from the point of tangency without intersecting the curve at another point.

Next, estimate the derivatives, which are the slopes of these tangent lines:
For \( x = -1 \): The slope of the tangent is approximately 3, so \( f'(-1) \approx 3 \)
For \( x = 0 \): The tangent is horizontal, with a slope of 0, so \( f'(0)= 0 \)
For \( x = 1 \): The slope is approximately 3, hence \( f'(1) \approx 3 \)
This step shows the rate of change of the function at specific points, providing an insightful understanding of its behavior.

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Most popular questions from this chapter

The economic order quantity, EOQ, is used in cost accounting to minimize the total cost, TC, to order and carry a firm's stock over the period of a year. The annual cost of placing orders, \(A C O\), is given by $$ \mathrm{ACO}=\frac{(\mathrm{ARU})(\mathrm{CO})}{\mathrm{EOQ}} $$ where $$ \begin{aligned} \mathrm{ARU} &=\text { annual required units } \\ \mathrm{CO} &=\text { cost per order } \end{aligned} $$ The annual carrying cost, ACC, is given by $$ \mathrm{ACC}=(\mathrm{CU})(\mathrm{CC}) \frac{(\mathrm{EOQ})}{2} $$ where $$ \begin{aligned} &\mathrm{CU}=\text { cost per unit } \\ &\mathrm{CC}=\text { carrying } \mathrm{cost} \end{aligned} $$ and \((E O Q) / 2\) provides an estimate of the average number of units in stock at any given time of the year. Assuming that \(\mathrm{ARU}, \mathrm{CO}, \mathrm{CU}\) and \(\mathrm{CC}\) are all constant, show that the total cost $$ \mathrm{TC}=\mathrm{ACO}+\mathrm{ACC} $$ is minimized when $$ \mathrm{EOQ}=\sqrt{\frac{2(\mathrm{ARU})(\mathrm{CO})}{(\mathrm{CU})(\mathrm{CC})}} $$

Consider the supply equation $$ Q=4+0.1 P^{2} $$ (a) Write down an expression for \(\mathrm{dQ} / \mathrm{d} P\). (b) Show that the supply equation can be rearranged as $$ P=\sqrt{(10 Q-40)} $$ Differentiate this to find an expression for \(\mathrm{d} P / \mathrm{d} Q\). (c) Use your answers to parts (a) and (b) to verify that $$ \frac{\mathrm{d} Q}{\mathrm{~d} P}=\frac{1}{\mathrm{~d} P / \mathrm{d} Q} $$ (d) Calculate the elasticity of supply at the point \(Q=14\).

Given the demand equation $$ P=-Q^{2}-10 Q+150 $$ find the price elasticity of demand when \(Q=4\). Estimate the percentage change in price needed to increase demand by \(10 \%\).

If the average cost function of a good is. $$ \mathrm{AC}=\frac{15}{Q}+2 Q+9 $$ find an expression for TC. What are the fixed costs in this case? Write down an expression for the marginal cost function.

If the savings function is given by $$ S=0.02 Y^{2}-Y+100 $$ calculate the values of \(M P S\) and \(M P C\) when \(Y=40\). Give a brief interpretation of these results.
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