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Chapter 7: Problem 1

In Problems 1-5, find the linear approximation of \(f(x)\) at \(x=0\). $$ f(x)=e^{2 x} $$

Short Answer

Expert verified
The linear approximation of \( f(x) = e^{2x} \) at \( x=0 \) is \( L(x) = 1 + 2x \).

Step by step solution

01

Understand Linear Approximation

The linear approximation of a function at a point is the tangent line to the function at that point. The equation of this tangent line can be given by: \[ L(x) = f(a) + f'(a)(x-a) \]where \( a \) is the point of approximation, \( f(a) \) is the function value at \( a \), and \( f'(a) \) is the derivative of the function evaluated at \( a \). In our problem, \( a = 0 \).
02

Determine the Function Value at x=0

Substitute \( x = 0 \) into the function to find \( f(0) \): \[ f(0) = e^{2 \cdot 0} = e^0 = 1 \]Thus, \( f(0) = 1 \).
03

Differentiate the Function

Find the derivative of the function \( f(x) = e^{2x} \) with respect to \( x \): \[ f'(x) = \frac{d}{dx} e^{2x} = 2e^{2x} \]So the derivative is \( f'(x) = 2e^{2x} \).
04

Evaluate the Derivative at x=0

Substitute \( x = 0 \) into the derivative to find \( f'(0) \): \[ f'(0) = 2e^{2\cdot0} = 2e^0 = 2 \]Thus, \( f'(0) = 2 \).
05

Construct the Linear Approximation

Using the linear approximation formula \( L(x) = f(a) + f'(a)(x-a) \), substitute \( f(0) = 1 \), \( f'(0) = 2 \), and \( a = 0 \): \[ L(x) = 1 + 2(x - 0) = 1 + 2x \]So the linear approximation is \( L(x) = 1 + 2x \).

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Key Concepts

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

Tangent Line
The tangent line to a function at a particular point is a straight line that just touches the curve of the function without crossing it. This line has the same slope as the function at that specific point.
To visualize this, imagine a smooth curve representing the function graph. The tangent line is like a tiny segment of that curve, appearing straight.
The tangent line's equation is often used as a linear approximation for the function near the point of tangency, making it easier to work with around that point.
Derivative
The derivative of a function provides crucial information about the function's rate of change at any point. It tells us how the function's output will change in response to tiny changes in the input.
For any function, the derivative at a point represents the slope of the tangent line at that point. If you think of the function's graph as a mountain slope, the derivative is the measure of the steepness at each point.
In the context of function analysis, finding the derivative is the first step in determining how the function behaves near any particular point.
Function Approximation
Function approximation, particularly linear approximation, is a technique used to estimate the value of a function using simpler expressions. This approach is especially useful when dealing with complex functions.
Linear approximation uses the tangent line at a known point on a curve to predict the function's behavior in the vicinity of that point. This is done by crafting a linear equation that closely resembles the function around that area.
By employing linear approximation, intricate calculations can be simplified, making it a powerful tool in analysis and problem-solving for both engineering and basic calculus applications.

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Most popular questions from this chapter

a, b, and \(c\) are constants and \(g(x)\) is a continuous function whose derivative \(g^{\prime}(x)\) is also continuous. Use substitution to evaluate the indefinite integrals. $$ \int g^{\prime}(x) e^{-g(x)} d x $$

Use integration by parts to evaluate the integrals. $$ \int x \sec ^{2} x d x $$

Use substitution to evaluate the definite integrals. $$ \int_{2}^{3} \frac{2 x+3}{\left(x^{2}+3 x\right)^{3}} d x $$

Use integration by parts to verify the validity of the reduction formula $$ \int(\ln x)^{n} d x=x(\ln x)^{n}-n \int(\ln x)^{n-1} d x $$ (b) Apply the reduction formula in (a) repeatedly to compute $$ \int(\ln x)^{3} d x $$

Use integration by parts to evaluate the integrals. $$ \int 2 x \cos (3 x-1) d x $$
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