### Calculating the Approximate Derivative Using Given ValuesTo approximate the derivative at a given point using nearby function values, we can use the definition of the derivative. Given the function values:\[ f(-10) = -0.1 \]\[ f(-9.6) = 1.8 \]we can approximate the derivative \( f'(-10) \) using the difference quotient formula for the derivative:\[ f'(x) \approx \frac{f(x+h) - f(x)}{h} \]In this case:- \( x = -10 \)- \( x+h = -9.6 \) so \( h = -9.6 - (-10) = 0.4 \)Substitute these values into the formula:\[ f'(-10) \approx \frac{f(-9.6) - f(-10)}{-9.6 - (-10)} = \frac{1.8 - (-0.1)}{0.4} = \frac{1.8 + 0.1}{0.4} = \frac{1.9}{0.4} \approx 4.75 \]Therefore, the approximate value of \( f'(-10) \) is:\[ f'(-10) \approx 4.75 \]To better understand this approximation, it is important to visualize how nearby function values can help us estimate the rate of change or the slope of the tangent to the function at a point. This technique is particularly useful for functions that are not easily differentiable by standard methods, or when we only have discrete data points.#### Summary:Given values:- \( f(-10) = -0.1 \)- \( f(-9.6) = 1.8 \)Approximation:- \( f'(-10) \approx 4.75 \)

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Answered: Given that f(-10) = -0.1 and f(-9.6) =…

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Given that f(-10) = -0.1 and f(-9.6) = 1.8, approximate f'(-10). f'(-10) ~ ≈ 94

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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### Calculating the Approximate Derivative Using Given Values

To approximate the derivative at a given point using nearby function values, we can use the definition of the derivative. Given the function values:

\[ f(-10) = -0.1 \]
\[ f(-9.6) = 1.8 \]

we can approximate the derivative \( f'(-10) \) using the difference quotient formula for the derivative:

\[ f'(x) \approx \frac{f(x+h) - f(x)}{h} \]

In this case:
- \( x = -10 \)
- \( x+h = -9.6 \) so \( h = -9.6 - (-10) = 0.4 \)

Substitute these values into the formula:

\[ f'(-10) \approx \frac{f(-9.6) - f(-10)}{-9.6 - (-10)} = \frac{1.8 - (-0.1)}{0.4} = \frac{1.8 + 0.1}{0.4} = \frac{1.9}{0.4} \approx 4.75 \]

Therefore, the approximate value of \( f'(-10) \) is:

\[ f'(-10) \approx 4.75 \]

To better understand this approximation, it is important to visualize how nearby function values can help us estimate the rate of change or the slope of the tangent to the function at a point. This technique is particularly useful for functions that are not easily differentiable by standard methods, or when we only have discrete data points.

#### Summary:

Given values:
- \( f(-10) = -0.1 \)
- \( f(-9.6) = 1.8 \)

Approximation:
- \( f'(-10) \approx 4.75 \)

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