Problem 64 - University Calculus: Early Transcendentals

Question

Suppose that \(0

Step-by-Step Solution

Answer

Yes, since \(0 < f(1) - f(-1) < 1\), the results satisfy the given inequality.

Step 1: Understanding the Problem

We are given a function \( f \) where the derivative is positive and less than \( 1/2 \) for all \( x \). We need to show that \( f(-1) < f(1) < 2 + f(-1) \). This means \( f \) is strictly increasing but flat enough to be bounded according to the given inequality.

Step 2: Using the Fundamental Theorem of Calculus

Since \( f \) is differentiable, we know that \( f(1) - f(-1) = \int_{-1}^{1} f'(x) \, dx \). Therefore, we can use the properties of integrals to deduce bounds on \( f(1) \).

Step 3: Bounding the Integral

The inequality \( 0 < f'(x) < 1/2 \) for all \( x \) means that \( f'(x) \) is always positive but never more than \( 1/2 \). Integrating over the interval from \(-1\) to \(1\), we get:\[0 < \int_{-1}^{1} f'(x) \, dx < \int_{-1}^{1} \frac{1}{2} \, dx\]Since the length of the interval \([-1, 1]\) is \(2\), this evaluates to:\[0 < f(1) - f(-1) < 1\]

Step 4: Reformulating the Inequality

From the integral evaluation, we derived \( 0 < f(1) - f(-1) < 1 \), meaning \( f(-1) < f(1) < 1 + f(-1) \). We originally need to show \( f(-1) < f(1) < 2 + f(-1) \). The inequality \( f(1) < 1 + f(-1) \) satisfies this condition because any \( x \) less than \( 1 \) is also less than \( 2 \).

Step 5: Conclusion

The condition \( 0 < f(1) - f(-1) < 1 \) leads us to conclude \( f(-1) < f(1) < 1 + f(-1) \), which fits the general condition \( f(-1) < f(1) < 2 + f(-1) \). This shows that the function \( f(x) \) fulfills the requirement of the problem.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the world of derivatives and integrals, two of the main operations in calculus. It asserts that integration and differentiation are inverse processes.

For any function that is continuous on a closed interval \([a, b]\), the theorem can be understood in two parts:

• First, if a function \(F\) has a continuous derivative \(f\) on \([a, b]\), then \[F(b) - F(a) = \int_{a}^{b} f(x) \, dx\] shows that the definite integral of \(f\) from \(a\) to \(b\) is the net variation of an antiderivative \(F\) across the interval.
• Second, it establishes that the derivative of the integral of a function \(f\) from a to \(x\) is the function itself, i.e., \[F'(x) = f(x)\] if \ F(x) = \int_{a}^{x} f(t) \, dt.\

In solving inequalities where derivatives are bounded, like in our exercise, this theorem empowers us to shift from a derivative perspective to an integral view, to understand how function values vary over an interval.

Differentiable functions

A function is termed "differentiable" at a point if it has a derivative there, implying the function's graph has a defined tangent at that point. Differentiability is a stronger condition than continuity; a differentiable function is always continuous, but not every continuous function is differentiable.

When a function is differentiable over an interval, like in the problem where \(f'(x)\) is positive and less than \(1/2\), it indicates the function is smooth and provides a local linear approximation everywhere in the range.

• This means the function has no sharp corners or cusps within the interval.
• The derivative's bounds \(0 < f'(x) < 1/2\) tell us that the function's slope is positive but gentle, which ensures that \(f\) increases steadily across the interval.

Understanding differentiability helps us utilize the Fundamental Theorem of Calculus. It aids in connecting derivative bounds with function growth over specific intervals.

Inequalities in Calculus

Inequalities in calculus often involve bounding the behavior of functions, derivatives, or integrals. They are a key component in understanding how functions behave under certain conditions.

In our exercise, we're given an inequality for the derivative \(f'(x)\), allowing us to draw conclusions about the original function.

• Given \(0 < f'(x) < 1/2\), it means \(f(x)\) is increasing but at a controlled rate. This inequality restricts both the steepness and the total growth of \(f\) over any interval, such as \([-1, 1]\).
• By integrating \(f'(x)\), we can establish bounds for \(f(-1) < f(1) < 1 + f(-1)\), ensuring \(f(1)\) does not surpass \(f(-1) + 1\).

This approach confirms that inequalities serve as tools to predict and explain the behavior, slope, and growth of functions, creating a more complete picture of function dynamics. By understanding how functions behave within certain bounds, one leverages inequalities to make precise predictions about changes across their entire domain.