Problem 113
Question
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.
Step-by-Step Solution
Verified Answer
The curve is \(y = \ln(x)\) and is concave down for \(x > 0\).
1Step 1: Understand the Derivative
We are given that the first derivative is \(f'(x) = \frac{1}{x}\). This suggests that the slope of the tangent to the curve increases as \(x\) increases from 1 onwards. Since \(f'(x) > 0\) for \(x > 0\), the function is increasing for all positive \(x\).
2Step 2: Find the Antiderivative
Integrate \(f'(x)\) to find \(f(x)\). The antiderivative of \(\frac{1}{x}\) is \(\ln|x|\). Therefore, a general form for \(f(x)\) is \(f(x) = \ln(x) + C\).
3Step 3: Solve for the Constant
Use the given point \(f(1) = 0\) to find the constant \(C\). Substituting \(x = 1\) into \(f(x) = \ln(x) + C\), we have \(0 = \ln(1) + C = 0 + C\). Thus, \(C = 0\).
4Step 4: Write the Specific Function
From the previous step, we find \(f(x) = \ln(x)\). This is the specific function that satisfies all given conditions.
5Step 5: Analyze the Concavity
Determine the concavity by examining the second derivative \(f''(x)\). Given \(f(x) = \ln(x)\), the second derivative \(f''(x) = -\frac{1}{x^2}\). Since \(-\frac{1}{x^2}\) is always negative for \(x > 0\), the function is concave down for \(x > 0\).
Key Concepts
Understanding DerivativesExploring AntiderivativesUnderstanding Concavity
Understanding Derivatives
Derivatives are a fundamental concept in calculus, used to determine the rate at which a function is changing. Essentially, a derivative provides the slope of the tangent line to a function at any given point. In simple terms, it tells us how a function is increasing or decreasing.
In our exercise, we are given that the derivative of the function is \(f'(x) = \frac{1}{x}\). This implies that the slope of the function changes with \(x\). Specifically, for \(x > 0\), \(f'(x) > 0\), which indicates that the function is continually increasing for positive values of \(x\).
Key points about derivatives:
In our exercise, we are given that the derivative of the function is \(f'(x) = \frac{1}{x}\). This implies that the slope of the function changes with \(x\). Specifically, for \(x > 0\), \(f'(x) > 0\), which indicates that the function is continually increasing for positive values of \(x\).
Key points about derivatives:
- They help in finding rates of change.
- A positive derivative means the function is increasing.
- A negative derivative means the function is decreasing.
- A zero derivative implies a constant function.
Exploring Antiderivatives
Antiderivatives, also known as indefinite integrals, are the reverse operation of derivatives. Finding an antiderivative involves determining the original function given its derivative. This process is called integration.
From the exercise, the derivative given is \(f'(x) = \frac{1}{x}\). Its antiderivative is \(\int \frac{1}{x} \, dx = \ln|x| + C\), where \(C\) is a constant. This result helps us find a formula for the function \(f(x)\).
To find the specific function, we use the condition \(f(1) = 0\). Substituting \(x = 1\) into \(f(x) = \ln(x) + C\), we get \(0 = \ln(1) + C\). Since \(\ln(1) = 0\), it follows that \(C = 0\). Thus, the function is \(f(x) = \ln(x)\).
Importance of antiderivatives:
From the exercise, the derivative given is \(f'(x) = \frac{1}{x}\). Its antiderivative is \(\int \frac{1}{x} \, dx = \ln|x| + C\), where \(C\) is a constant. This result helps us find a formula for the function \(f(x)\).
To find the specific function, we use the condition \(f(1) = 0\). Substituting \(x = 1\) into \(f(x) = \ln(x) + C\), we get \(0 = \ln(1) + C\). Since \(\ln(1) = 0\), it follows that \(C = 0\). Thus, the function is \(f(x) = \ln(x)\).
Importance of antiderivatives:
- They allow us to reconstruct the original function from its rate of change.
- Indefinite integrals include an arbitrary constant \(C\) because differentiation of a constant is zero.
- Provide solutions to differential equations.
Understanding Concavity
Concavity describes the curvature of a graph. It tells us whether a function bends upwards or downwards. To determine concavity, we use the second derivative of the function.
For our specific function \(f(x) = \ln(x)\), we first find the second derivative \(f''(x) = -\frac{1}{x^2}\). Because \(-\frac{1}{x^2} < 0\) for \(x > 0\), the function is concave down across this interval.
Key points about concavity:
For our specific function \(f(x) = \ln(x)\), we first find the second derivative \(f''(x) = -\frac{1}{x^2}\). Because \(-\frac{1}{x^2} < 0\) for \(x > 0\), the function is concave down across this interval.
Key points about concavity:
- If the second derivative \(f''(x) > 0\), the function is concave up.
- If \(f''(x) < 0\), the function is concave down.
- Concave up functions resemble a bowl facing upwards.
- Concave down functions resemble a bowl facing downwards.
Other exercises in this chapter
Problem 112
Suppose the derivative of the function \(y=f(x)\) is $$y^{\prime}=(x-1)^{2}(x-2)(x-4)$$ At what points, if any, does the graph of \(f\) have a local minimum, lo
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Can anything be said about the graph of a function \(y=f(x)\) that has a continuous second derivative that is never zero? Give reasons for your answer.
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