/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Sketch the graph of a function \... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketch the graph of a function \(f\) that has the following properties. (a) has a continuous first derivative; (b) is decreasing and concave up for \(x<3\) (c) has an extremum at \((3,1)\); (d) is increasing and concave up for \(36\).

Short Answer

Expert verified
Sketch a smooth curve with local minima at \((3,1)\) and maxima at \((6,7)\), changing concavity at \((5,4)\).

Step by step solution

01

Identify the function properties

Analyze each given property one by one. Property (a) implies that the function's graph is smooth with no sharp turns. Properties (b) and (d) collectively suggest the function is decreasing before \(x=3\) and increasing after \(x=3\). An extremum at \((3,1)\) in (c) suggests a local minimum. The inflection point at \((5,4)\) in (e) indicates that the concavity changes from up to down. Property (f) means the function increases until \(x=6\), where it reaches another extremum as specified in (g). The function then decreases for \(x>6\) according to (h).
02

Graph sketch from properties for x

From property (b), sketch the graph such that for \(x<3\), the function is decreasing and concave up. This indicates a downward-sloping curve whose slope becomes less negative as \(x\) approaches 3.
03

Include the extremum at (3,1)

Mark the point \((3,1)\) on the graph. Since this is a local minimum, the graph should reach its lowest point here and start increasing immediately after \(x=3\).
04

Graph behavior for 3

In this interval, based on property (d), sketch the graph as increasing and remaining concave up, leading up to \(x=5\), where the concavity will change.
05

Identify inflection at (5,4)

Plot \((5,4)\) as the inflection point where the concavity changes. To the left of (5,4), the graph is concave up, and to the right, it starts being concave down.
06

Graph for 5

From property (f), sketch the graph as increasing but now concave down until \(x=6\). The curvature now bows downward but remains rising until 6.
07

Extremum at (6,7)

Plot \((6,7)\) on the graph to show that it is a local maximum. The graph should peak here before it starts decreasing for \(x>6\).
08

Graph for x>6

Following property (h), illustrate the graph decreasing and remaining concave down after \(x=6\). This indicates a downward-sloping curve that bows downward.

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Key Concepts

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

Continuity of Derivatives
In calculus, continuity of derivatives is an important aspect when sketching graphs. A derivative being continuous means that the graph of the function is smooth and lacks sharp turns or corners. This has implications on how we perceive and draw the graph. For the function given in the exercise, having a continuous first derivative ensures that we can transition smoothly through each behavior stipulated by the properties of this function.
This continuous nature guarantees that the function maintains a certain flow in changes happening across the x-values. In essence, the absence of abrupt changes or gaps in the slopes means we need not worry about jumps in our sketching process. Knowing the derivative is continuous supports detailed understanding and careful sketching of the function.
Concavity in Functions
Concavity describes how the direction of the curve is shaped, and it’s tied to the second derivative of a function. If you imagine a bowl, concave up would look like the interior of the bowl, and concave down would appear like the exterior. Understanding concavity helps identify the natural tendencies of the curve, such as where it might bend or change direction.
In the given problem, the function is said to be concave up when it is both decreasing and increasing at different x-values before hitting the inflection point at 5. Beyond this, it turns concave down where the increasing section before the maximum becomes less steep. Identifying where these changes occur can be crucial in graphing the overall behavior of the function accurately, leading up to changes in concave nature.
Local Extrema
Local extrema refer to points on a graph where the function reaches a local maximum or minimum. These points are particularly important because they indicate where the function temporarily stops and changes its direction. In the exercise, two points are identified: \(3,1\), a local minimum, and \(6,7\), a local maximum.
At the local minimum \(3,1\), the function reaches the lowest point in that neighborhood, meaning it decreases before this point and starts to increase afterward. Conversely, at the local maximum \(6,7\), the curve peaks, rising sharply earlier and beginning to drop after this point. Locating these extrema can help in identifying key features and pivotal points that shape the overall graph.
Inflection Points
Inflection points are fascinating since they mark where the function changes concavity. This is where the second derivative changes sign, implying an altered slope behavior. In the provided graph, point \(5,4\) serves as such a point. Before reaching \(5,4\), the function is concave up, where the graph curves upwards, but immediately after it transitions to concave down, curving towards downward slope.
This change at \(5,4\) is critical. It implies that the bending of the function shifts, introducing new characteristics to the graph’s form on either side of this point. Recognizing and marking inflection points helps build the graph's general appearance and a detailed understanding of the function's underlying behaviors.

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Most popular questions from this chapter

In Problems \(1-4\), show that the indicated function is a solution of the given differential equation; that is, substitute the indicated function for \(y\) to see that it produces an equality. $$ \frac{d y}{d x}+\frac{x}{y}=0 ; y=\sqrt{1-x^{2}} $$

A car is stationary at a toll booth. Eighteen minutes later at a point 20 miles down the road the car is clocked at 60 miles per hour. Sketch a possible graph of \(v\) versus \(t\). Sketch a possible graph of the distance traveled \(s\) against \(t .\) Use the Mean Value Theorem to show that the car must have exceeded the 60 mile per hour speed limit at some time after leaving the toll booth, but before the car was clocked at 60 miles per hour.

Consider the equation \(x=2\left(x-x^{2}\right)=g(x)\). (a) Sketch the graph of \(y=x\) and \(y=g(x)\) using the same coordinate system, and thereby approximately locate the positive root of \(x=g(x)\). (b) Try solving the equation by the Fixed-Point Algorithm starting with \(x_{1}=0.7\). (c) Solve the equation algebraically.

\mathrm{\\{} E X P L ~ B e s i d e s ~ p r o v i d i n g ~ a n ~ e a s y ~ w a y ~ t o ~ d i f f e r e n t i a t e ~ p r o d u c t s , ~ l o g . ~ arithmic differentiation also provides a measure of the relative or fractional rate of change, defined as $y^{\prime} / y Prove that if the relative rate of change is a negative constant then the function must represent exponential decay.

First find the general solution (involving a constant \(\mathrm{C}\) ) for the given differential equation. Then find the particular solution that satisfies the indicated condition. $$ \frac{d y}{d x}=(2 x+1)^{4} ; y=6 \text { at } x=0 $$

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