Problem 22
Question
Estimate the instantaneous rate of change of the function \(f(x)=x \ln x\) at \(x=1\) and at \(x=2 .\) What do these values suggest about the concavity of the graph between \(\mathrm{J}\) and \(2 ?\)
Step-by-Step Solution
Verified Answer
The graph is concave up between \( x=1 \) and \( x=2 \).
1Step 1: Recognize the Need for a Derivative
To estimate the instantaneous rate of change of the function at specific points, we need to find the derivative \( f'(x) \) of the function \( f(x) = x \ln x \).
2Step 2: Use the Product Rule for Differentiation
Since \( f(x) = x \ln x \) is a product of two functions, we will use the product rule: \( (uv)' = u'v + uv' \). Here, let \( u = x \) and \( v = \ln x \). The derivative of \( u \) is \( u' = 1 \), and the derivative of \( v \) is \( v' = \frac{1}{x} \).
3Step 3: Differentiate the Function
Applying the product rule: \[ f'(x) = (1)(\ln x) + (x)(\frac{1}{x}) = \ln x + 1. \]Thus, the derivative is \( f'(x) = \ln x + 1 \).
4Step 4: Plug in x = 1 to Find f'(1)
Substitute \( x = 1 \) into the derivative: \[ f'(1) = \ln 1 + 1 = 0 + 1 = 1. \]
5Step 5: Plug in x = 2 to Find f'(2)
Substitute \( x = 2 \) into the derivative: \[ f'(2) = \ln 2 + 1 \approx 0.693 + 1 = 1.693. \]
6Step 6: Analyze the Rates for Concavity
The values \( f'(1) = 1 \) and \( f'(2) \approx 1.693 \) indicate that the rate of change is increasing. This suggests the graph is concave up between \( x = 1 \) and \( x = 2 \).
Key Concepts
Instantaneous Rate of ChangeProduct RuleConcavity Analysis
Instantaneous Rate of Change
The instantaneous rate of change of a function at a particular point gives us the slope of the tangent line at that point. Imagine driving a car and looking at your speedometer at a specific moment; that's similar to what the instantaneous rate of change tells you about a function. It shows the exact rate the function's output changes with respect to changes in input, at that particular point.
To find this rate, we use derivatives. Derivatives tell us the rate at which the function value is changing for any infinitesimally small change in input. Hence, for the function \(f(x) = x \ln x\), its instantaneous rate of change at a point \(x\) is given by its derivative, \(f'(x) = \ln x + 1\).
To find this rate, we use derivatives. Derivatives tell us the rate at which the function value is changing for any infinitesimally small change in input. Hence, for the function \(f(x) = x \ln x\), its instantaneous rate of change at a point \(x\) is given by its derivative, \(f'(x) = \ln x + 1\).
- At \(x = 1\), we computed \(f'(1) = 1\), meaning at \(x = 1\), the function is increasing at a rate of 1.
- At \(x = 2\), \(f'(2) \approx 1.693\), suggesting that at \(x = 2\), the function is increasing faster, at a rate of approximately 1.693.
Product Rule
The product rule is a vital tool in differentiation used to handle situations when a function is composed of two differentiable functions multiplied together. If you have a function \(f(x)\) that is the product of two functions \(u(x)\) and \(v(x)\), i.e., \(f(x) = u(x)v(x)\), the product rule allows you to find the derivative. It is stated as \((uv)' = u'v + uv'\).
In our scenario, for \(f(x) = x \ln x\):
\[f'(x) = 1(\ln x) + x\left(\frac{1}{x}\right) = \ln x + 1.\]
This simplifies the process of finding the derivative of a function that involves multiplication, making it easier to understand and apply.
In our scenario, for \(f(x) = x \ln x\):
- Let \(u = x\), which means \(u' = 1\).
- Let \(v = \ln x\), so \(v' = \frac{1}{x}\).
\[f'(x) = 1(\ln x) + x\left(\frac{1}{x}\right) = \ln x + 1.\]
This simplifies the process of finding the derivative of a function that involves multiplication, making it easier to understand and apply.
Concavity Analysis
Concavity refers to the direction which a curve bends. A graph is concave up when its slope is increasing, and concave down when its slope is decreasing. The second derivative of a function, \(f''(x)\), can be used to determine concavity:
When analyzing between \(x = 1\) and \(x = 2\):
- If \(f''(x) > 0\), the function is concave up.
- If \(f''(x) < 0\), the function is concave down.
When analyzing between \(x = 1\) and \(x = 2\):
- The derivative goes from \(f'(1) = 1\) to \(f'(2) \approx 1.693\).
- This shows the slope is becoming steeper, indicating that the function is concave up in this interval.
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