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Chapter 4: Problem 11

Sketch the graph of a function on the inter val \(0 \leq x \leq 10\) with the given properties. Has local and global minimum at \(x=3\), local and global maximum at \(x=8\).

Short Answer

Expert verified
Draw a graph decreasing to \(x=3\), increasing to \(x=8\), and decreasing to \(x=10\).

Step by step solution

01

Identifying Critical Points

We are given two critical points: a local and global minimum at \(x = 3\) and a local and global maximum at \(x = 8\). These points will determine the shape of the graph in this interval.
02

Checking Intervals

Between \(x=0\) and \(x=3\), the function must be decreasing. From \(x=3\) to \(x=8\), the function must be increasing to go from the minimum to the maximum. Finally, from \(x=8\) to \(x=10\), the function must decrease again.
03

Sketching the Graph

Begin drawing from \(x=0\) with a generally decreasing trend until \(x=3\). At \(x=3\), plot the lowest point of the graph. Then, sketch an increasing trend from this point up to \(x=8\), where the graph reaches its peak. Lastly, draw a downward trend from \(x=8\) to \(x=10\).
04

Finalizing the Shape

Ensure the graph at \(x=3\) has a "U" shape (indicating a minimum) and an upside-down "U" shape at \(x=8\) (indicating a maximum). Verify the end behavior: the left end at \(x=0\) should be higher than \(x=3\), and the right end at \(x=10\) should be lower than \(x=8\).

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Key Concepts

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

Critical Points
Critical points are special places on a graph where the function's derivative is either zero or undefined. These points often signal where a graph might change direction or have a peak or trough. For our exercise, the given critical points are at
  • \(x = 3\) - for a local and global minimum
  • \(x = 8\) - for a local and global maximum
At these points, the function's slope is flat, meaning the rate of change is zero. Identifying critical points is vital because they help in sketching the general shape of a graph. When solving for critical points, look for where the first derivative of the function equals zero, or where the function does not exist.
Local Minimum
A local minimum is a point on a graph where the function has a lower value than at any nearby points. It looks like the bottom of a "U" shape. In the exercise, the local minimum occurs at \(x = 3\). Before reaching a local minimum, the function typically decreases. After this point, it increases. This change suggests that the slope was negative before the local minimum and becomes positive afterward. Checking this transition helps confirm a local minimum's location. Local minimums are essential for understanding the smaller trends within a function. They tell you where the function bottoms out locally, offering insights into the function's behavior in confined intervals.
Local Maximum
A local maximum is where a function peaks at a higher point than its surroundings, resembling the top of an inverted "U" shape. For this problem, The local maximum is at \(x = 8\). Before reaching this local maximum, the function increases. After this point, it decreases. Visualizing this shift helps you recognize where the local maximum is located. The slope transitions from positive to negative at a local maximum, indicating this peak in the graph. Local maximums are helpful for identifying where smaller parts of the graph reach their highest points, giving you insight into the function's local behavior.
Global Minimum
The global minimum is the single lowest point on the entire graph. It represents the absolute smallest y-value the function achieves over its whole domain. In this exercise, the global minimum happens at \(x = 3\). This is the same point as the local minimum because there isn't a lower point anywhere on the graph. It's essential to distinguish between local and global minimums. The global minimum provides the lowest point over the entire range, which, in some problems, might differ from any local minimum if the domain expands. The global minimum will always be the lowest point when comparing all possible outputs of the function.
Global Maximum
The global maximum is the highest point on the whole graph. It is where the function reaches its maximum value across the entire domain. In the exercise, this occurs at \(x = 8\). This point also serves as the local maximum. The function's value at this point is higher than that of any other point in the interval from \(x = 0\) to \(x = 10\). Knowing the global maximum is essential because it tells you the highest possible output the function can achieve. Like the global minimum, the global maximum stands out on the graph when considering all potential points on the function. Recognizing this point ensures you understand the function's ultimate peak over the desired range.

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Most popular questions from this chapter

A company can produce and sell \(f(L)\) tons of a product per month using \(L\) hours of labor per month. The wage of the workers is \(w\) dollars per hour, and the finished product sells for \(p\) dollars per ton. (a) The function \(f(L)\) is the company's production function. Give the units of \(f(L) .\) What is the practical significance of \(f(1000)=400 ?\) (b) The derivative \(f^{\prime}(L)\) is the company's marginal product of labor. Give the units of \(f^{\prime}(L) .\) What is the practical significance of \(f^{\prime}(1000)=2 ?\) (c) The real wage of the workers is the quantity of product that can be bought with one hour's wages. Show that the real wage is \(w / p\) tons per hour. (d) Show that the monthly profit of the company is $$ \pi(L)=p f(L)-w L $$ (e) Show that when operating at maximum profit, the company's marginal product of labor equals the real wage: $$ f^{\prime}(L)=\frac{w}{p} $$

Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. \(f(x)=x^{4}-4 x^{3}+10\)

A warehouse selling cement has to decide how often and in what quantities to reorder. It is cheaper, on average, to place large orders, because this reduces the ordering cost per unit. On the other hand, larger orders mean higher storage costs. The warehouse always reorders cement in the same quantity, \(q\). The total weekly cost, \(C\), of ordering and storage is given by \(C=\frac{a}{q}+b q, \quad\) where \(a, b\) are positive constants. (a) Which of the terms, \(a / q\) and \(b q\), represents the ordering cost and which represents the storage cost? (b) What value of \(q\) gives the minimum total cost?

Find the exact global maximum and minimum values of the function. The domain is all real numbers unless otherwise specified. $$ g(x)=4 x-x^{2}-5 $$

If \(q\) is the quantity of chicken demanded as a function of the price \(p\) of beef, the cross-price elasticity of demand for chicken with respect to the price of beef is defined as \(E_{\text {cross }}=|p / q \cdot d q / d p|\). What does \(E_{\text {cross }}\) tell you about the sensitivity of the quantity of chicken bought to changes in the price of beef?
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