Problem 56
Question
Use the given information to estimate \(f^{\prime}(c)\) at the given point \(c\) $$ f(4)=5.7 \text { and } f(4.1)=6.2, c=4.1 $$
Step-by-Step Solution
Verified Answer
The estimated value of \( f'(c) \) at \( c = 4.1 \) is 5.
1Step 1: Identify the Difference Quotient Formula
The derivative of a function at a point can be approximated as the difference quotient: \( f'(c) \approx \frac{f(x+h) - f(x)}{h} \), where \( h \) is a small increment.
2Step 2: Assign Values to the Formula
Here, we are given \( f(4) = 5.7 \) and \( f(4.1) = 6.2 \). We want to estimate \( f'(4.1) \). Assign \( x = 4 \) and \( h = 0.1 \), so the formula becomes \( f'(4.1) \approx \frac{f(4.1) - f(4)}{0.1} \).
3Step 3: Plug in the Values
Substitute the given values: \( f'(4.1) \approx \frac{6.2 - 5.7}{0.1} \).
4Step 4: Calculate the Result
Compute the numerator: \( 6.2 - 5.7 = 0.5 \). Then divide by \( 0.1 \): \( f'(4.1) \approx \frac{0.5}{0.1} = 5 \).
Key Concepts
Difference QuotientDerivative EstimationFunction Approximation
Difference Quotient
In differential calculus, the concept of a difference quotient is pivotal for understanding how derivatives work. A difference quotient helps us approximate the rate of change of a function over a small interval. Imagine you are examining how a car's speed changes over a very short distance; the difference quotient gives us that insight by measuring changes between two points.
- It is expressed as: \( \frac{f(x+h) - f(x)}{h} \), where \( f(x) \) represents the function value at a point \( x \), and \( h \) is a very small increment.
- This approach helps us "zoom in" on the curve of a function, by looking at the slope between two closely spaced points.
- As \( h \) approaches zero, the difference quotient becomes the derivative, represented by \( f'(x) \).
Derivative Estimation
Estimating a derivative involves calculating the rate at which the function's value changes at a particular point. This is important because direct measurement of the derivative can be difficult or impossible in many real-world applications. We rely on approximation techniques like using a difference quotient.
- In the context of the given exercise, we do not have \( f'(x) \) directly. Instead, we approximate it using known function values.
- By substituting \( f(4.1) \) and \( f(4) \) into the difference quotient formula, we derive an estimate for \( f'(4.1) \).
- This is achieved by calculating \( \frac{6.2 - 5.7}{0.1} \), which gives us the estimated derivative.
Function Approximation
Function approximation is a crucial technique when dealing with continuously changing systems. It allows us to model and predict behaviors by looking at a small section of the function and expanding our understanding from there.
- In practical terms, this involves estimating or "guessing" the behavior of a function at points where direct measurement isn't feasible.
- In the exercise, although we don't know \( f'(4.1) \) exactly, we can approximate it by evaluating the slope between two close function points: \( f(4) \) and \( f(4.1) \).
- Approximations are especially useful in scientific and engineering contexts, where complex models must be simplified to predict outcomes efficiently.
Other exercises in this chapter
Problem 56
In Exercises \(56-59\), evaluate the derivative \(f^{\prime}\) of the given function \(f\) in two ways. First, apply the Chain Rule to \(f(x)\) without simplify
View solution Problem 56
Compute \(f^{\prime}(\mathrm{c})\) for the given \(f\) and \(c\). \(f(x)=\left(x^{5}+x-1\right)\left(2 x^{2}-3\right)^{2}, c=-1\)
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What is the rate of change of the area of a square with respect to its side length when the side length is 8 centimeters?
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Let \(f(x)=x /\left(1+x^{2}\right)^{2} .\) Use the identity $$ \frac{x /\left(1+x^{2}\right)^{2}-c /\left(1+c^{2}\right)^{2}}{x-c}=\frac{1-c x^{3}-c^{2} x^{2}-2
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