Problem 19
Question
Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ $$ f(x)=\log x \text { at } a=1 $$
Step-by-Step Solution
Verified Answer
The linear approximation of \( \log x \) at \( x = 1 \) is \( L(x) = x - 1 \).
1Step 1: Identify given functions and values
We are given the function \( f(x) = \log x \) and need to approximate it at \( a = 1 \). We also know the formula for linear approximation is \( L(x) = f(a) + f'(a)(x-a) \).
2Step 2: Compute \( f(a) \)
First, substitute \( a = 1 \) into the function to find \( f(1) \).\[ f(1) = \log 1 \]Since the logarithm of 1 is 0, we have \( f(1) = 0 \).
3Step 3: Find \( f'(x) \), the derivative of \( f(x) \)
The derivative of \( f(x) = \log x \) with respect to \( x \) is \( f'(x) = \frac{1}{x} \).
4Step 4: Compute \( f'(a) \)
Now substitute \( a = 1 \) into the derivative to find \( f'(1) \).\[ f'(1) = \frac{1}{1} = 1 \]
5Step 5: Plug into the linear approximation formula
Using the values \( f(a) = 0 \) and \( f'(a) = 1 \), substitute them into the linear approximation formula: \[ L(x) = 0 + 1 \cdot (x-1) = x-1 \]
Key Concepts
Understanding the DerivativeExploring Logarithmic FunctionsEffective Approximation Methods
Understanding the Derivative
The derivative is a fundamental concept in calculus, used to determine the rate at which a function changes. In the context of our exercise, it helps us find the slope of the tangent line to the curve of the function at a specific point. This slope is crucial for linear approximation.
To find the derivative of a function, one generally looks at the limit of the average rate of change as the interval approaches zero. For the function \( f(x) = \log x \), the derivative is obtained using the rule that the derivative of \( \log x \) with respect to \( x \) is \( \frac{1}{x} \), indicating that the rate of change of \( \log x \) diminishes as \( x \) increases.
This derivative tells us that as each unit increase in \( x \) occurs, \( \log x \) increases by an amount decreasing proportionally to the value of \( x \). It plays a significant role when performing approximations or analyzing the behavior of logarithmic functions around a specific point.
To find the derivative of a function, one generally looks at the limit of the average rate of change as the interval approaches zero. For the function \( f(x) = \log x \), the derivative is obtained using the rule that the derivative of \( \log x \) with respect to \( x \) is \( \frac{1}{x} \), indicating that the rate of change of \( \log x \) diminishes as \( x \) increases.
This derivative tells us that as each unit increase in \( x \) occurs, \( \log x \) increases by an amount decreasing proportionally to the value of \( x \). It plays a significant role when performing approximations or analyzing the behavior of logarithmic functions around a specific point.
Exploring Logarithmic Functions
Logarithmic functions, like \( f(x) = \log x \), are essential in various fields such as science, engineering, and finance. They describe phenomena like exponential growth or decay, where growth rates reduce as the variable increases.
The function \( \log x \) is the inverse of the exponential function. Simply put, if \( y = \log x \), then \( x = 10^y \) for common logarithms. One key property of logs is that they transform multiplicative relationships into additive ones, which is useful for simplifying complex equations.
At \( x = 1 \), the value of \( \log x \) is 0, which means multiplying any number by 1 doesn't change it, aligns with our understanding of multiplication. This property simplifies calculations, making logarithmic functions beneficial when estimating large or complex calculations in a manageable way.
The function \( \log x \) is the inverse of the exponential function. Simply put, if \( y = \log x \), then \( x = 10^y \) for common logarithms. One key property of logs is that they transform multiplicative relationships into additive ones, which is useful for simplifying complex equations.
At \( x = 1 \), the value of \( \log x \) is 0, which means multiplying any number by 1 doesn't change it, aligns with our understanding of multiplication. This property simplifies calculations, making logarithmic functions beneficial when estimating large or complex calculations in a manageable way.
Effective Approximation Methods
Approximation methods like linear approximation allow us to estimate values of complex functions close to a particular point with simple linear expressions. This is especially helpful in scenarios where calculating the exact values is cumbersome or impossible.
Linear approximation involves using the tangent line at a specific point to approximate the function. The formula \( L(x)=f(a)+f'(a)(x-a) \) is applied to create a linear function. For the function \( \log x \) at \( a = 1 \), the line \( L(x) = x - 1 \) gives an easy way to estimate values near \( x = 1 \).
Such methods reduce the complexity of problems and are widely used in optimization and computational algorithms where simplicity and speed are crucial. Understanding approximation techniques is vital for fields such as physics, economics, and machine learning, where exact precision is often less critical than computational efficiency.
Linear approximation involves using the tangent line at a specific point to approximate the function. The formula \( L(x)=f(a)+f'(a)(x-a) \) is applied to create a linear function. For the function \( \log x \) at \( a = 1 \), the line \( L(x) = x - 1 \) gives an easy way to estimate values near \( x = 1 \).
Such methods reduce the complexity of problems and are widely used in optimization and computational algorithms where simplicity and speed are crucial. Understanding approximation techniques is vital for fields such as physics, economics, and machine learning, where exact precision is often less critical than computational efficiency.
Other exercises in this chapter
Problem 18
Apply the product rule to find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. $$ f(x)=(1-2 x)(1+2 x), \text { at } x=2 $$
View solution Problem 18
Differentiate the functions given in Problems with respect to the independent variable. $$ f(x)=\frac{x}{e}+e^{2} x+e $$
View solution Problem 19
Compute \(f(c+h)-f(c)\) at the indicated point. $$ f(x)=\sqrt{x} ; c=4 $$
View solution Problem 19
Differentiate the functions with respect to the independent variable. \(f(r)=\left(r^{2}-r\right)^{3}\left(r+3 r^{3}\right)^{-4}\)
View solution