Problem 41 - Calculus - AP Edition

Question

Find the equation of the tangent line to the graph of \(y=f(x)\) at \(x=x_{0}\). $$f(x)=\ln x ; x_{0}=e^{-1}$$

Step-by-Step Solution

Answer

The tangent line equation is \( y = ex - 2 \).

Step 1: Find the Derivative

To find the equation of the tangent line, we first need to determine the derivative of the function, which gives us the slope of the tangent line. The function is given by \( f(x) = \ln x \). The derivative of \( \ln x \) is \( \frac{1}{x} \). Thus, \( f'(x) = \frac{1}{x} \).

Step 2: Evaluate the Derivative at the Given Point

Next, we evaluate the derivative at the point \( x = e^{-1} \) to find the slope of the tangent line. Substitute \( x_{0} = e^{-1} \) into \( f'(x) \): \[ f'(e^{-1}) = \frac{1}{e^{-1}} = e. \]Hence, the slope of the tangent line at \( x = e^{-1} \) is \( e \).

Step 3: Find the Function Value at the Given Point

To find the equation of the tangent line, we also need the y-coordinate of the point of tangency. Substitute \( x_{0} = e^{-1} \) into the original function \( f(x) = \ln x \):\[ f(e^{-1}) = \ln(e^{-1}) = -1. \]Thus, the point of tangency is \( (e^{-1}, -1) \).

Step 4: Write the Equation of the Tangent Line

Using the point-slope form of a line, \( y - y_{1} = m(x - x_{1}) \), where \((x_{1}, y_{1})\) is the point of tangency \((e^{-1}, -1)\), and \( m = e \) is the slope:\[ y - (-1) = e(x - e^{-1})\]Simplifying, we get:\[y + 1 = ex - 1\]Rearranging the equation, the tangent line is:\[y = ex - 2.\]

Step 5: Conclusion

The equation of the tangent line to the function \( y = \ln x \) at \( x = e^{-1} \) is \( y = ex - 2 \).

Key Concepts

Tangent Line

When you hear the term "tangent line," think of a line that just "touches" the curve at a specific point. It doesn’t cut across the curve, it merely grazes it at exactly one spot.
To know the equation of this line, two things are must-haves: the slope of the tangent at the desired point and the point itself.

• The slope tells us how steep or flat the line is at the point.
• The point is where the tangent makes contact with the curve.

Remember that the tangent line is a powerful concept in calculus because it approximates the function near the point of tangency. This is often useful in various applications, from physics to economics, where you want to estimate changes close to a certain point.

Derivative

The derivative is a fundamental concept in calculus used to determine how a function is changing at any given point. For a given function, its derivative function gives you a new function that describes the original function's rate of change or slope.
Taking the derivative involves various rules, depending on the type of function you're dealing with.

• For the natural logarithm function, denoted as \( \ln x \), the derivative is \( \frac{1}{x} \).

This derivative expression \( \frac{1}{x} \) tells us how quickly or slowly the natural logarithm function \( \ln x \) is rising or falling at any point \( x \). When calculated at a specific value like \( x = e^{-1} \), it gives us the slope of the tangent line at that point.

Natural Logarithm

The natural logarithm, written as \( \ln x \), is a special logarithmic function. This logarithm is "natural" because it is based on Euler's number, \( e \), approximately equal to 2.718.
Why use \( \ln x \)? One reason is that it simplifies many mathematical expressions, especially those involving continuous growth or decay processes, such as interest rates or populations.

• The natural logarithm is the inverse of the exponential function \( e^x \).
• Key property: \( \ln e = 1 \), because \( e^1 = e \).

The function \( \ln x \) is defined only for \( x > 0 \). As \( x \) gets larger, \( \ln x \) increases, but at a decreasing rate. This behavior is reflected in its derivative, \( \frac{1}{x} \), which shows that the slope decreases as \( x \) increases. This makes \( \ln x \) an interesting function to explore when learning about growth rates and changes.