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Chapter 4: Problem 103

For \(x>0,\) sketch a curve \(y=f(x)\) that has \(f(1)=0\) and \(f^{\prime}(x)=1 / x .\) Can anything be said about the concavity of such a curve? Give reasons for your answer.

Short Answer

Expert verified
The curve is \( y = \ln x \), increasing and concave down for \( x > 0 \).

Step by step solution

01

Identify the Derivative Function

We are given that the derivative of the function is \( f'(x) = \frac{1}{x} \). This means that at any point \( x \), the slope of the tangent to the curve is equal to \( \frac{1}{x} \). This derivative is positive for all \( x > 0 \), indicating that \( f(x) \) is increasing on this interval.
02

Find the General Form of the Function

To find \( f(x) \), integrate \( f'(x) = \frac{1}{x} \). The integral of \( \frac{1}{x} \) is \( \ln|x| + C \). Since \( x > 0 \), this simplifies to \( f(x) = \ln x + C \).
03

Use the Given Condition to Find the Constant

We know that \( f(1) = 0 \). Substituting \( x=1 \) in the equation \( f(x) = \ln x + C \), we get \( 0 = \ln 1 + C \). Since \( \ln 1 = 0 \), we find that \( C = 0 \). Thus, the function is \( f(x) = \ln x \).
04

Evaluate the Second Derivative for Concavity

Calculate the second derivative \( f''(x) \) to assess concavity. Since \( f'(x) = \frac{1}{x} \), differentiate again to find \( f''(x) = -\frac{1}{x^2} \). The second derivative \( f''(x) \) is negative for \( x > 0 \), indicating the curve is concave down on \( x > 0 \).
05

Sketch the Curve and Note the Concavity

The curve \( y = \ln x \) is increasing and concave down for \( x > 0 \), passing through the point \( (1, 0) \). Draw the graph of \( y = \ln x \), which rises slowly to the right and gets infinitely close to the y-axis from the right as it approaches x = 0.

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Key Concepts

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

Derivative
Derivatives help us understand how a function changes. It's like measuring the steepness of a hill at any point. For the function given, the derivative is defined as \( f'(x) = \frac{1}{x} \). This derivative tells us that, for every positive \( x \), the slope of the function is \( \frac{1}{x} \).
  • Since \( \frac{1}{x} \) is always positive when \( x > 0 \), the function is increasing.
  • An increasing function climbs up; it never goes down as \( x \) increases.
Remember, the derivative is crucial to predict how rapidly or slowly a function changes at any given point.Analyzing the derivative helps us understand the behavior of the function and its graph.
Integral
Integrals are essentially the opposite of derivatives. They help us find the original function when we only know how it changes.Here, we're given \( f'(x) = \frac{1}{x} \). To find the function \( f(x) \), we integrate this derivative:
  • The integral of \( \frac{1}{x} \) is \( \ln|x| + C \).
  • Because \( x > 0 \), this simplifies to \( \ln x + C \).
  • We use the condition \( f(1) = 0 \) to solve for \( C \), finding \( C = 0 \).
  • This means our function is \( f(x) = \ln x \).
Integrals allow us to trace a function's path and a clue into its original form before differentiating.
Concavity
Concavity tells us about how a function curves and bends. If a graph looks like a cup, opening up or down, concavity shows us this shape.To assess concavity, we look at the second derivative:
  • Start with the first derivative: \( f'(x) = \frac{1}{x} \).
  • Differentiate again to get the second derivative: \( f''(x) = -\frac{1}{x^2} \).
  • Because \( f''(x) < 0 \) for \( x > 0 \), the graph of \( f(x) \) is concave down.
This concave down nature means the graph will curve like an upside-down bowl. Concavity gives insight into the nature of the graph's curvature.
Function Graph
The graph of a function visually represents how the function behaves.For \( f(x) = \ln x \), the graph has some distinct characteristics:
  • It passes through the point \( (1, 0) \).
  • As \( x \) increases, the graph slowly rises to the right. Since \( x > 0 \), our interest is rightward.
  • The graph will never touch or cross the y-axis, but it gets closer and closer. It's termed asymptotic to the y-axis as \( x \rightarrow 0^+ \).
  • Given the concave down property, the graph arcs down, never forming a cup upwards.
Sketching and understanding a function's graph lets you predict its behavior intuitively.

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Most popular questions from this chapter

Liftoff from Earth A rocket lifts off the surface of Earth with a constant acceleration of \(20 \mathrm{m} / \mathrm{s}^{2} .\) How fast will the rocket be going 1 min later?

Motion along a coordinate line A particle moves on a coordinate line with acceleration \(a=d^{2} s / d t^{2}=15 \sqrt{t}-(3 / \sqrt{t})\) subject to the conditions that \(d s / d t=4\) and \(s=0\) when \(t=1\) Find a. the velocity \(v=d s / d t\) in terms of \(t\) b. the position \(s\) in terms of \(t\).

Solve the initial value problems. $$\frac{d v}{d t}=\frac{1}{2} \sec t \tan t, \quad v(0)=1$$

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int \frac{t \sqrt{t}+\sqrt{t}}{t^{2}} d t$$

Right, or wrong? Say which for each formula and give a brief reason for each answer. a. \(\int x \sin x d x=\frac{x^{2}}{2} \sin x+C\) b. \(\int x \sin x d x=-x \cos x+C\) c. \(\int x \sin x d x=-x \cos x+\sin x+C\)
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