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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 \(2 \leq x \leq 4\)

Short Answer

Expert verified
The linearization of \(f(x) = \log_{3}(x)\) at \(x = 3\) is \(L(x) = 1 + 0.30(x - 3)\). Graph both \(f(x)\) and \(L(x)\) for the interval \(0 \leq x \leq 8\) and \(2 \leq x \leq 4\).

Step by step solution

01

Understand the Problem

We need to find the linearization of the function \( f(x) = \log_{3}(x) \) at \( x = 3 \). Linearization is an approximation of the function around a point using its tangent line.
02

Find the Derivative

To find the linearization, we first need the derivative of the function, \( f(x) = \log_{3}(x) \). Remember that the derivative of \( \log_{b}(x) \) is \( \frac{1}{x \ln(b)} \). Thus, the derivative is \( f'(x) = \frac{1}{x \ln(3)} \).
03

Evaluate the Function and Derivative at x=3

Evaluate \( f(x) \) and \( f'(x) \) at \( x = 3 \). For \( f(x) \):\( f(3) = \log_{3}(3) = 1 \).For \( f'(x) \):\( f'(3) = \frac{1}{3 \ln(3)} \).
04

Write the Linearization Formula

The linearization \( L(x) \) of a function at a point \( x = a \) is given by:\[ L(x) = f(a) + f'(a)(x - a) \]Substitute \( a = 3 \), \( f(3) = 1 \), and \( f'(3) = \frac{1}{3 \ln(3)} \) into this formula:\[ L(x) = 1 + \left(\frac{1}{3 \ln(3)}\right)(x - 3) \]
05

Simplify and Round Coefficients

Calculate \( \ln(3) \) approximately as 1.0986. Thus,\[ L(x) = 1 + \frac{1}{3 \times 1.0986}(x - 3) \]Simplify the coefficient of \( (x - 3) \):\[ \frac{1}{3 \cdot 1.0986} \approx 0.303 \]So, the linearization simplifies to:\[ L(x) = 1 + 0.303(x - 3) \]Rounded to two decimal places:\[ L(x) = 1 + 0.30(x - 3) \]
06

Plot the Function and Linearization

Plot \( f(x) = \log_{3}(x) \) and its linearization \( L(x) = 1 + 0.30(x - 3) \) within the windows \( 0 \leq x \leq 8 \) and \( 2 \leq x \leq 4 \).Check that the tangent line \( L(x) \) closely approximates \( f(x) \) around \( x = 3 \). In graphing software, use these bounds to visualize how well the linearization matches the actual function within these intervals.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of Logarithmic Functions
Understanding derivatives is a crucial step in working with functions like logarithms. When dealing with the function \( f(x) = \log_{3}(x) \), the derivative indicates the rate at which the function changes at a certain point. For logarithmic functions \( \log_{b}(x) \), the general formula for the derivative is \( \frac{1}{x \ln(b)} \).

This means, for \( f(x) = \log_{3}(x) \), the derivative \( f'(x) \) becomes \( \frac{1}{x \ln(3)} \). The natural logarithm, denoted as \( \ln \), is used here to account for the base change from base \( b \) to the natural logarithm's base \( e \).

At \( x = 3 \), to find how steep the curve is, we substitute \( x = 3 \) into the derivative. Thus, \( f'(3) = \frac{1}{3 \ln(3)} \). By plugging in an approximate value of \( \ln(3) \) as 1.0986, the computation becomes manageable and results in \( 0.303 \) approximately. This value approximates the slope at that specific point on the curve.
Tangent Line Approximation
Tangent line approximation, often referred to as linearization, helps us approximate the values of functions near a point. This is particularly handy when the function is complicated but behaves almost linearly at small intervals. Linearizing a function involves using its value and its derivative at a specific point to form a simple linear function.

The formula for linearization \( L(x) \) at a point \( x = a \) is derived as: \[ L(x) = f(a) + f'(a)(x - a) \]
In this case, for \( f(x) = \log_{3}(x) \) and \( a = 3 \), we've calculated \( f(3) = 1 \) and \( f'(3) = 0.303 \). By plugging these into our formula, we get:
\[ L(x) = 1 + 0.303(x - 3) \]
We round the coefficient \( 0.303 \) to \( 0.30 \) for a nicely simplified result.

Thus, our tangent line approximation at \( x = 3 \) is \( L(x) = 1 + 0.30(x - 3) \), which makes it easier to compute values close to \( x = 3 \). This linear function closely matches the logarithmic curve at this point and gives a straight-line approximation.
Graphing Functions
Graphing the function \( f(x) = \log_{3}(x) \) alongside its linear approximation presents a clear picture of how linearization works.
In our specific scenario, we plot both the function and its linearization \( L(x) = 1 + 0.30(x - 3) \) in the range \( 0 \leq x \leq 8 \) and \( 2 \leq x \leq 4 \).

This range covers a portion of the curve where the approximation holds true. For effective visualization:
  • The logarithmic function shows a gently increasing curve since logarithms increase slowly as x grows.
  • The tangent line, meanwhile, is straight, capturing the behavior accurately near \( x = 3 \).
By using graphing tools, one can observe that near \( x = 3 \), the tangent line and the curve are nearly indistinguishable. As you move further from this point, the difference becomes noticeable, showcasing the limitation of linear approximations.
Concluding, graphing not only aids in understanding function behavior at specific points but also displays the practicality and bounds of linearization.

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