Problem 68 - University Calculus: Early Transcendentals

Question

a. Find the linearization of \(f(x)=\log _{3} x\) at \(x=3 .\) Then round its coefficients to two decimal places. b. Graph the linearization and function together in the window \(0 \leq x \leq 8\) and \(2 \leq x \leq 4\)

Step-by-Step Solution

Answer

Linearization: \(L(x) = 0.30x + 0.09\). Graph both functions over the specified intervals to compare.

Step 1: Understand the Function and Linearization Point

The function given is \(f(x) = \log_{3} x\). We're tasked with finding the linearization at \(x = 3\). Linearization approximates a function around a specific point using the tangent line equation.

Step 2: Find the Derivative of the Function

To perform linearization, we need the derivative of the function at the point. The derivative of \(\log_{3} x\) is \(f'(x) = \frac{1}{x \ln 3}\).

Step 3: Calculate the Derivative at the Linearization Point

Plug \(x = 3\) into the derivative:\[f'(3) = \frac{1}{3 \ln 3}.\]Calculating, \(\ln 3 \approx 1.0986\), so \(f'(3) \approx 0.303\) after rounding to three decimal places.

Step 4: Find the Function Value at the Linearization Point

Evaluate the original function at \(x = 3\):\[f(3) = \log_3 3 = 1.\]

Step 5: Write the Equation of the Tangent Line

The equation of the tangent line (linearization) is given by:\[L(x) = f(a) + f'(a)(x - a).\]Substitute \(a = 3\), \(f(3) = 1\), and \(f'(3) \approx 0.303\):\[L(x) = 1 + 0.303(x - 3).\]

Step 6: Simplify the Linearization Equation

Simplify the equation:\[L(x) = 1 + 0.303x - 0.909.\]This simplifies to:\[L(x) = 0.303x + 0.091.\] Round the coefficients to two decimal places: \(L(x) = 0.30x + 0.09\).

Step 7: Graph the Function and the Linearization

Graph both \(f(x) = \log_3 x\) and \(L(x) = 0.30x + 0.09\) over the intervals \(0 \leq x \leq 8\) and \(2 \leq x \leq 4\). The graphs will show \(L(x)\) closely approximating \(f(x)\) around \(x = 3\), but diverging as you move further away from this point.

Key Concepts

Linearization

When dealing with calculus, linearization is a strategy used to approximate complex functions with simpler linear ones. This is particularly helpful when calculations around particular points are needed. Essentially, it's like creating a tangent line that hugs the curve of the function at a specific point. For any function \( f(x) \), its linearization around a point \( x = a \) can be expressed as a straight line equation:

• \( L(x) = f(a) + f'(a)(x - a) \)

Here, \( a \) is the point where you're approximating the function with a tangent line, \( f(a) \) is the function's value at \( x = a \), and \( f'(a) \) is the derivative of the function at that point.
Linearization is useful for making predictions and understanding the behavior of functions near a specific value, simplifying the task of calculation around tricky points.

Derivative

The derivative of a function is a core concept in calculus. It represents the rate at which the function's value changes as its input changes. You can think of it as measuring how the "slope" of a function's graph behaves.
For a function \( f(x) \), the derivative \( f'(x) \) tells us the slope of the tangent line to the graph of the function at any given point. This not only gives us insight into whether the function is increasing or decreasing at that point but also allows us to make estimates about the function's values near that point.

• For example, the derivative of \( \log_{3} x \) is \( f'(x) = \frac{1}{x \ln 3} \).

This means that at any point \( x \) on the graph, the slope is \( \frac{1}{x \ln 3} \). When you substitute \( x = 3 \), this becomes \( \frac{1}{3 \ln 3} \), approximating the slope to 0.303. This piece of information is essential when creating the linear approximation or the tangent line at that point.

Tangent Line

The tangent line to a function at a specific point is like a flat road touching a curve at just one spot. It gives the best linear approximation to the function near that point.
The equation of a tangent line is derived using the value of the function and its derivative at the given point. For \( f(x) = \log_{3} x \), the tangent line at \( x = 3 \) is calculated as follows:

• The function value at \( x = 3 \) is \( f(3) = 1 \).
• The derivative, or slope, at this point is \( f'(3) \approx 0.303 \).

So, the equation of the tangent line or the linearization at \( x = 3 \) is \( L(x) = 1 + 0.303(x - 3) \), simplifying to \( L(x) = 0.303x - 0.909 \). When rounded, the linearization becomes \( L(x) = 0.30x + 0.09 \).
Graphically, this line will touch the curve of \( f(x) \) at exactly one point and will approximately "stay" with it near \( x = 3\), providing a visual understanding of how the function behaves at that point.