Problem 11 - Calculus: Early Transcendentals

Question

Suppose that \(f\) is a differentiable function such that \(f^{\prime}(x) \geq-3\) for all \(x\). What is the smallest possible value of \(f(4)\) if \(f(-1)=2 ?\)

Step-by-Step Solution

Answer

The smallest possible value of \( f(4) \) is \( -13 \).

Step 1: Understanding the Problem

We need to find the smallest possible value of \( f(4) \) given that \( f(-1) = 2 \) and the derivative \( f'(x) \geq -3 \) for all \( x \). The derivative tells us the rate at which \( f(x) \) is changing.

Step 2: Using the Mean Value Theorem

The Mean Value Theorem states that for a differentiable function \( f \) on the interval \([-1, 4]\), there exists a point \( c \) in \((-1, 4)\) such that \( f'(c) = \frac{f(4) - f(-1)}{4 - (-1)} \). Here, \( f'(c) \geq -3 \), so we use this information to find constraints on \( f(4) \).

Step 3: Setting up the Inequality

Substitute the known values into the inequality: \(\frac{f(4) - 2}{4 - (-1)} \geq -3\). Simplify this to get \( \frac{f(4) - 2}{5} \geq -3 \), which can be rewritten as \( f(4) - 2 \geq -15 \).

Step 4: Solving for \( f(4) \)

Simplify the inequality \( f(4) - 2 \geq -15 \) by adding 2 to both sides, yielding \( f(4) \geq -13 \). This implies that the smallest possible value for \( f(4) \) is \( -13 \).

Key Concepts

Differentiable Function

A differentiable function is a fundamental concept in calculus. It refers to a function that has a derivative at each point in its domain. This means the function's graph is smooth, without any sharp edges or cusps. Differentiability implies continuity, which means the function's behavior is predictable and regular. Here, we are dealing with a function \( f \) that is differentiable, which allows the application of the Mean Value Theorem. The theorem uses the existence of a derivative to draw conclusions about the function's values at certain points. Understanding differentiable functions helps students link the abstract concept of derivatives with concrete outcomes, like finding the smallest or largest possible values of \( f \) under given conditions.

Derivative Constraint

In this exercise, the derivative of \( f \) is constrained by the inequality \( f'(x) \geq -3 \) for all \( x \). This kind of constraint provides valuable information about the function's rate of change. Specifically, it tells us that while \( f \) may decrease, it cannot decrease faster than a rate of -3 units per unit of \( x \).

• If \( f'(x) \) were greater, the function could rise more quickly.
• If \( f'(x) \) were exactly -3, the function decreases at this maximum allowed rate.

Understanding derivative constraints allows us to predict the minimum or maximum values that \( f \) can reach, given its initial conditions and range of \( x \). This prediction is based on the worst-case scenario for the function's slope at any point in its domain.

Inequality Solving

In the solution, we use inequalities to express conditions set by the Mean Value Theorem. These reflect the potential growth or decrease within the range \([-1, 4]\) for our differentiable function \( f(x) \). When we substitute into the theorem, we get the inequality \( \frac{f(4) - 2}{5} \geq -3 \). This step is crucial as it sets up a range within which \( f(4) \) must fall.


By solving the inequality, we perform basic algebraic transformations:

• First, multiplying through by 5 to get \( f(4) - 2 \geq -15 \).
• Then, adding 2 to both sides to isolate \( f(4) \), resulting in \( f(4) \geq -13 \).

Solving inequalities helps students dissect a problem by translating conditions into mathematical expressions. By understanding this, one gains insights into how changes in one part of an equation affect the entire system, allowing us to pinpoint crucial values like the smallest possible \( f(4) \) in this case.