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Chapter 4: Problem 30

\(\begin{array}{l}{\text { Suppose } f(3)=2, f^{\prime}(3)=\frac{1}{2}, \text { and } f^{\prime}(x)>0 \text { and } f^{\prime \prime}(x)<0} \\ {\text { for all } x .}\end{array}\) $$\begin{array}{l}{\text { (a) Sketch a possible graph for f. }} \\ {\text { (b) How many solutions does the equation } f(x)=0 \text { have? }} \\ {\text { Why? }} \\ {\text { (c) Is it possible that } f^{\prime}(2)=\frac{1}{3} ? \text { Why? }}\end{array}$$

Short Answer

Expert verified
(a) Sketch: Concave down, increasing curve through \((3, 2)\). (b) One solution; \(f(x)\) crosses zero once. (c) Possible; no constraints against \(f'(2) = \frac{1}{3}\).

Step by step solution

01

Understand the Given Conditions

The function is such that \(f(3) = 2\), meaning the point \((3, 2)\) is on the graph. \(f'(3) = \frac{1}{2}\) shows the slope of the tangent at \(x = 3\) is positive. \(f'(x) > 0\) for all \(x\), so \(f(x)\) is strictly increasing. \(f''(x) < 0\) means the graph is concave down everywhere, implying the function is always bending downwards.
02

Sketch a Possible Graph for f

Start from a point slightly above the \(x\)-axis at \(x = 0\) (since it's positive and increasing) and pass through \((3, 2)\) with a positive slope. Ensure the graph is concave down at every point by approximating the curve's shape to a part of a downward-bending parabola or similar curve.
03

Determine Solutions to f(x) = 0

Since \(f(x)\) is strictly increasing and \(f(3) = 2\), it must cross the \(x\)-axis to increase to the point \((3, 2)\). However, due to strictly positive derivative, it can only cross once, moving from negative to positive before \(x = 3\). Therefore, it has exactly one real root.
04

Justify the Number of Solutions

Because \(f(x)\) is strictly increasing, if it starts below zero, it must cross the \(x\)-axis once to become positive and increase to \((3, 2)\). The strict positivity of \(f'(x)\) ensures no turning points where it could return to 0.
05

Examine possibility of f'(2) = 1/3

Given \(f'(x) > 0\) for all \(x\), this constant positive derivative can take any positive value, and \(\frac{1}{3}\) is positive. There's no condition preventing \(f'(2)\) from being \(\frac{1}{3}\) since it's a plausible tangent slope value considering the derivative's constraints.

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Key Concepts

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

Concavity
When discussing concavity in calculus, it refers to the direction in which a curve bends. A function can be either concave up or concave down.
If a function is concave up, the graph looks like an upward-opening cup, while a concave down function resembles a downward-opening bowl.
In our exercise, the function has your concavity always pointing downwards, meaning it is concave down throughout. This is indicated by the second derivative, \(f''(x) < 0\).
This results in the function graph bending downwards as you move from left to right.
  • This influences the shape of the graph significantly, making sure it doesn't form any peaks or valleys that go upwards.
  • In other words, the graph can never "turn upward" at any point.
Understanding concavity helps in identifying the overall shape and behavior of the graph of a function along its entire domain.
Increasing Function
An increasing function is one where, as you move to the right along the x-axis, the function's output values rise. It signifies that the function always climbs as you progress forward. This idea is represented by the first derivative.
In the given exercise, it tells us that \(f'(x) > 0\) for all \(x\), meaning that our function is strictly increasing everywhere on its domain.
  • No matter where you look on the graph, the overall trend is going upward.
  • It never dips or levels off at any point.
This increasing nature is a major contributor to understanding when and how a function might intersect the x-axis. An intersection would mean finding solutions to the equation \(f(x) = 0\). Every increase implies that after intersecting the x-axis once, there will be no return to zero.
Solutions of Equations
Solutions to the equation \(f(x) = 0\) are the x-values where the graph of the function intersects the x-axis. For the current exercise, we know \(f(x)\) is strictly increasing, so it won't touch or cross the x-axis more than once.
Given the function passes through the point \((3, 2)\), it must have met, or will meet, the x-axis once before reaching this point, as it moves upward from some negative starting point.
  • This guarantees exactly one real root.
  • Strictly increasing nature and positive derivative ensure the function gets more and more positive as \(x\) increases.
Finding how many solutions is crucial as it guides us in understanding how the function behaves along its graph and the number of intersections at the x-axis.
Derivative
The derivative in calculus serves as a measure of how a function changes at any point. The first derivative, \(f'(x)\), gives the slope of the tangent to the graph at any given value of \(x\).
This helps to determine whether the function is increasing or decreasing at that point.
  • In our problem, \(f'(3) = \frac{1}{2}\) shows a positive slope, indicating an upward direction of the graph at \(x = 3\).
  • The positivity of \(f'(x)\) across all \(x\) indicates a non-stop increase.
The second derivative, \(f''(x)\), reflects concavity. Because \(f''(x) < 0\), the graph is bending downwards. Understanding the role of both first and second derivatives allows us to sketch the possible shape and track changes within the function's curve.

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Most popular questions from this chapter

Use a computer algebra system to graph \(f\) and to find \(f^{\prime}\) and \(f^{\prime \prime} .\) Use graphs of these derivatives to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points of \(f\). \(f(x)=\sqrt{x+5 \sin x}, \quad x \leqslant 20\)

\(5-8\) Use Newton's method with the specified initial approxima- tion \(x_{1}\) to find \(x_{3},\) the third approximation to the root of the given equation. (Give your answer to four decimal places.) \(\frac{1}{3} x^{3}+\frac{1}{2} x^{2}+3=0, \quad x_{1}=-3\)

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