Problem 27 Use the Mean Value Theorem to sh... [FREE SOLUTION]

Chapter 3: Problem 27

Use the Mean Value Theorem to show that \(s=1 / t\) decreases on any interval over which it is defined.

Short Answer

Expert verified

The function \( s=1/t \) decreases on intervals where it is defined because its derivative is negative.

Step by step solution

Understanding the Function

Let the function be defined as \( f(t) = \frac{1}{t} \). This function is defined for any interval where \( t > 0 \) or \( t < 0 \). We'll consider \( t > 0 \) for this explanation.

Calculate the Derivative of the Function

The derivative of \( f(t) = \frac{1}{t} \) is found using the power rule. Recall that \( \frac{1}{t} = t^{-1} \). Thus, the derivative \( f'(t) = \frac{d}{dt}(t^{-1}) = -t^{-2} = -\frac{1}{t^2} \).

Analyze the Sign of the Derivative

The derivative \( f'(t) = -\frac{1}{t^2} \) is negative for all \( t > 0 \) and \( t < 0 \). A negative derivative implies that the function is decreasing on the interval where it is defined.

Apply the Mean Value Theorem (MVT)

The Mean Value Theorem states that if a function \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a point \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). Since \( f(t) = \frac{1}{t} \) is continuous and differentiable for \( t > 0 \) or \( t < 0 \), MVT applies.

Conclusion

Since the derivative \( f'(t) \) is negative for \( t > 0 \) (and similarly for \( t < 0 \)), it means that \( \frac{f(b) - f(a)}{b - a} < 0 \). By the Mean Value Theorem, \( f(t) = \frac{1}{t} \) is guaranteed to be decreasing on any interval where it is defined.

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Key Concepts

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

Understanding Derivatives

In mathematics, a derivative is a measure of how a function changes as its input changes. Specifically, it tells us the rate at which a function's output changes in relation to changes in its input. Think of it as the function's "instantaneous rate of change" or "slope" at a particular point.

For a function defined as \( f(t) = \frac{1}{t} \), its derivative is represented as \( f'(t) \).
Derivatives are calculated using various rules, such as the power rule, which in this case gives us \( f'(t) = -t^{-2} \) or \(-\frac{1}{t^2} \).
This derivative is negative for all \( t eq 0 \), which means the function's slope is always decreasing in such intervals.

By understanding the sign of the derivative, we can draw conclusions about the behavior of the function itself. A negative derivative indicates that, as \( t \) increases or decreases, the function value \( f(t) \) itself decreases.

Concept of Continuous Functions

A function is said to be continuous if you can draw its graph without lifting your pen from the paper. This informal definition captures the essence of what continuity means: no breaks, jumps, or holes in the graph.

The formal definition is that a function \( f(t) \) is continuous at a point \( t = a \) if \( \lim_{t \to a} f(t) = f(a) \).
For the function \( f(t) = \frac{1}{t} \), it is continuous everywhere except at \( t = 0 \).
This means as long as \( t \) stays away from zero, the function will behave smoothly.

Being continuous is critical for applying the Mean Value Theorem, which requires a function to be continuous on a closed interval \([a, b]\) and differentiable on \((a, b)\). Thus, \( f(t) = \frac{1}{t} \) is an ideal candidate outside of \( t = 0 \).

Understanding Decreasing Functions

When we say a function is decreasing over an interval, it means that as the input variable increases, the output variable decreases. This is directly related to the sign of the derivative.

If the derivative \( f'(t) \) is negative over an interval, the function is decreasing in that interval.
For \( f(t) = \frac{1}{t} \), since \( f'(t) = -\frac{1}{t^2} \) is always negative for \( t > 0 \), it means \( f(t) \) is decreasing when \( t \) is positive.
A similar argument can be made for intervals where \( t < 0 \).

Using this information coupled with the Mean Value Theorem, we are assured that \( f(t) = \frac{1}{t} \) consistently decreases wherever it is defined. Thus, understanding and analyzing the derivative is key to verifying the function's decreasing nature.

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91影视

Use the Mean Value Theorem to show that \(s=1 / t\) decreases on any interval over which it is defined.

Short Answer

Step by step solution

Understanding the Function

Calculate the Derivative of the Function

Analyze the Sign of the Derivative

Apply the Mean Value Theorem (MVT)

Conclusion

Key Concepts

Understanding Derivatives

Concept of Continuous Functions

Understanding Decreasing Functions

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