Problem 103
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 \( y = \ln x \), increasing and concave down for \( x > 0 \).
1Step 1: 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.
2Step 2: 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 \( f(x) = \ln x + C \).
3Step 3: 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 \( f(x) = \ln x \).
4Step 4: 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 \( x > 0 \).
5Step 5: 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.
Key Concepts
DerivativeIntegralConcavityFunction Graph
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.
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 \( f(x) = \ln x \).
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.
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.
Other exercises in this chapter
Problem 102
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