Problem 70 - Contemporary Precalculus

Question

Find the average rate of change of the function. $$g(x)=x-\ln x, \text { as } x \text { goes from } .5 \text { to } 1$$

Step-by-Step Solution

Answer

Answer: The average rate of change is approximately $$1.61372$$.

Step 1: Find the function values at the endpoints

To find the function values at the endpoints of the interval [0.5, 1], we need to plug 0.5 and 1 into the function $$g(x) = x - \ln x$$. $$g(0.5) = 0.5 - \ln(0.5)$$ $$g(1) = 1 - \ln(1)$$ Evaluate these expressions to get the values of the function at each point.

Step 2: Evaluate the function values

Now we will calculate the values of $$g(0.5)$$ and $$g(1)$$. $$g(0.5) = 0.5 - \ln(0.5) \approx 0.19314$$ $$g(1) = 1 - \ln(1) = 1$$ We have $$g(0.5) \approx 0.19314$$ and $$g(1) = 1$$.

Step 3: Calculate the difference in function values

Next, we need to find the difference in function values between the two points. This is done by subtracting the function value at 0.5 from the function value at 1. $$\Delta g = g(1) - g(0.5) = 1 - 0.19314 \approx 0.80686$$ We have $$\Delta g \approx 0.80686$$.

Step 4: Calculate the difference in x-values

Now, we need to find the difference in x-values between the two points. This can be found by subtracting 0.5 from 1. $$\Delta x = 1 - 0.5 = 0.5$$ We have $$\Delta x = 0.5$$.

Step 5: Calculate the average rate of change

Finally, we can find the average rate of change of the function $$g(x) = x - \ln x$$ over the interval [0.5, 1] by dividing the difference in function values by the difference in x-values. Average Rate of Change = $$\frac{\Delta g}{\Delta x} \approx \frac{0.80686}{0.5} \approx 1.61372$$ Thus, the average rate of change of the function $$g(x) = x - \ln x$$ as x goes from 0.5 to 1 is approximately $$1.61372$$.

Key Concepts

Function Evaluation

Function evaluation is a process where we substitute a number for the variable in a function to determine the result. For our given function, \( g(x) = x - \ln x \), we evaluate it at the endpoints of the interval [0.5, 1]. This means we find the values of \( g(x) \) when \( x \) is 0.5 and again when \( x \) is 1.

• For \( x = 0.5 \), the function becomes \( g(0.5) = 0.5 - \ln(0.5) \).
• For \( x = 1 \), the function simplifies to \( g(1) = 1 - \ln(1) \).

Substituting 0.5 and 1 into the function gives us exact numerical values, which we then use to determine any changes or trends. This step is crucial for further calculations like average rate of change, as it provides the specific endpoints values needed.

Logarithmic Functions

Logarithmic functions are mathematical expressions that represent logarithms, which are the inverse operation to exponentiation. In essence, the logarithm of a number is the power to which the base must be raised to obtain that number. For the function \( g(x) = x - \ln x \), the logarithmic component is \( \ln x \). Here are a few key points about logarithms:

• The natural logarithm \( \ln x \) uses the base \( e \), where \( e \approx 2.71828 \).
• \( \ln 1 = 0 \), because any number raised to the power of 0 is 1.
• \( \ln x \) can provide negative values when \( 0 < x < 1 \), since our base \( e \) must be raised to a negative power to reach values less than 1.

Understanding how logarithmic functions behave helps us in simplifying expressions and solving calculus problems involving rates of change.

Calculus

Calculus is a branch of mathematics that studies how things change. One of its main concepts is the rate of change, which is essential when analyzing functions like \( g(x) = x - \ln x \). The average rate of change gives us an idea of how the function behaves over a specific interval, which, in this case, is from 0.5 to 1. Here's a breakdown of how we determine the average rate of change:

• First, calculate the change in function values: \( \Delta g = g(1) - g(0.5) \).
• Next, calculate the change in x-values: \( \Delta x = 1 - 0.5 \).
• The average rate of change is then \( \frac{\Delta g}{\Delta x} \), finding how much the function value changes per unit change in \( x \).

This rate helps in modeling real-world scenarios where understanding how functions change over intervals is crucial to predicting behavior and outcomes.