Chapter 4: Problem 49
Given \(f(x)=e^{k x}\), approximate \(f(h)\), where \(h\) is near zero, using a tangent-line approximation. \(f(h) \approx\) (A) \(k\) (B) \(k h\) (C) \(1+k\) (D) \(1+k h\)
Short Answer
Expert verified
The answer is (D) \(1 + kh\).
Step by step solution
01
Understand the Task
We are asked to approximate the function \(f(x) = e^{kx}\) near \(h = 0\) using a tangent-line approximation. This technique is also known as linearization, where we use the derivative to approximate a function.
02
Setup the Tangent Line Equation
The general form of the tangent-line approximation is \(f(x) \approx f(a) + f'(a) \cdot (x - a)\). Here, \(a = 0\) since we are interested in \(h\) near zero. We need \(f(0)\) and \(f'(0)\).
03
Evaluate the Function at \(a = 0\)
Calculate \(f(0) = e^{k \cdot 0} = e^0 = 1\). So, \(f(0) = 1\).
04
Find the Derivative of the Function
The derivative of \(f(x) = e^{kx}\) is computed using the chain rule. Therefore, \(f'(x) = k \cdot e^{kx}\).
05
Evaluate the Derivative at \(a = 0\)
Calculate \(f'(0) = k \cdot e^{k \cdot 0} = k \cdot e^0 = k\).
06
Apply the Tangent Line Approximation Formula
Substitute the values into the linear approximation formula: \[ f(h) \approx f(0) + f'(0) \cdot (h - 0) = 1 + k \cdot h \]
07
Determine the Best Matching Option
The tangent-line approximation expression \(1 + kh\) matches with option (D). Thus, the correct approximation when \(h\) is near zero is option (D).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Tangent-line approximation
The tangent-line approximation is a powerful technique in calculus used to estimate the value of a function at a point close to a known value, typically where the curve changes smoothly. Imagine you have a bendy curve, and you want to simplify it with a straight line so that small sections of the curve can be easily understood. That's what the tangent-line approximation does.
It uses the tangent at a particular point on the curve to approximate the curve nearby. This is especially useful when dealing with complex functions as it simplifies calculations. The approximation is based on a formula which leverages point values and the function's derivative:
It uses the tangent at a particular point on the curve to approximate the curve nearby. This is especially useful when dealing with complex functions as it simplifies calculations. The approximation is based on a formula which leverages point values and the function's derivative:
- The equation is: \[ f(x) \approx f(a) + f'(a) \cdot (x-a) \]
- It's mostly used when the point \(a\) and the new point \(x\) are closely situated.
Derivative
The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. Imagine pressing the accelerator pedal in a car; the derivative is akin to how your speed increases. Mathematically, the derivative gives us the slope or gradient of a function at any given point.
For a function like \(f(x) = e^{kx}\), calculating its derivative involves some rules:
For a function like \(f(x) = e^{kx}\), calculating its derivative involves some rules:
- **Chain Rule**: If you have a composite function, this rule helps find derivatives efficiently.
- For \(f(x) = e^{kx}\), the derivative \(f'(x) = k \cdot e^{kx}\) utilizes both the nature of exponential functions and the chain rule.
Linearization
Linearization is a technique closely linked with tangent-line approximation. It's essentially taking a curvy, complicated function and making it easy to understand by transforming it into a straight line. Think of it like taking a twisty mountain road and plotting out just the first few gentle curves, instead of the entire route.
This approach helps when you want to predict behavior or solve problems near a given point. Linearization simplifies the work by creating a linear model using the first derivative. This model predicts changes in the function value as \(x\) changes slightly:
This approach helps when you want to predict behavior or solve problems near a given point. Linearization simplifies the work by creating a linear model using the first derivative. This model predicts changes in the function value as \(x\) changes slightly:
- The linearized equation: \[ L(x) = f(a) + f'(a) \cdot (x-a) \]
- This equation is useful for approximations, especially when \(x\) is close to \(a\).