Problem 71 - Thomas Calculus

Question

Let \(f\) be differentiable at every value of \(x\) and suppose that \(f(1)=1,\) that \(f^{\prime}<0\) on \((-\infty, 1),\) and that \(f^{\prime}>0\) on \((1, \infty).\) \begin{equation}\begin{array}{l}{\text { a. Show that } f(x) \geq 1 \text { for all } x}. \\ {\text { b. Must } f^{\prime}(1)=0 ? \text { Explain. }}\end{array}\end{equation}

Step-by-Step Solution

Answer

a. Yes, \(f(x) \ge 1\) for all \(x\). b. Yes, \(f'(1)=0\) for differentiability at \(x=1\).

Step 1: Analyze the Derivative Behavior

First, identify how the function behaves based on the information given about its derivative. Since \( f' < 0 \) on \((-\infty, 1)\), \( f \) is decreasing in this interval. Conversely, since \( f' > 0 \) on \((1, \, \infty)\), \( f \) is increasing in this interval.

Step 2: Evaluate at Critical Point

Given that \( f(1) = 1 \), the critical point at \( x = 1 \) is where the behavior of \( f \) changes from decreasing to increasing. Therefore, \( x = 1 \) could be a local minimum for \( f \).

Step 3: Deduce the Value of \( f(x) \)

Since \( f(x) \) decreases to the left of \( x = 1 \) and increases to the right, the minimum value near \( x = 1 \) suggests that for all \( x \), \( f(x) \ge 1 \). This is because \( f(x) \) can only decrease to values \( \ge 1 \) from the left and increase from 1.

Step 4: Determine \( f'(1) \)

The information does not explicitly state that \( f'(1) = 0 \), but intuitively, there must be a transition between decreasing and increasing behavior. According to the first derivative test and differentiability at every \( x \), \( f'(1) \) must equal 0 for a local minima to exist, as \( f \) is continuous.

Key Concepts

Differentiability

Differentiability is a key concept in calculus that tells us about the smoothness of a function. A function is differentiable at a point if it has a derivative there, meaning we can draw a tangent line at that point without any sharp turns or cusps.
For the function in this exercise, it is stated that the function is differentiable at every value of \(x\). This assures us that it is smooth for all values and possesses a derivative everywhere.
Differentiability implies continuity, but the reverse is not always true. If you can take the derivative of a function at a point, you know it is continuous there as well.

Derivatives

Derivatives are central to understanding the behavior of functions. They tell us the rate at which a function changes as its input changes. The derivative of a function at a point can be thought of as the slope of the tangent line to the function's graph at that point.
In the given exercise, the derivative \(f'\) tells us how \(f\) behaves across different intervals:

• On \((-\infty, 1)\), \(f'(x) < 0\) indicating \(f\) is decreasing.
• On \((1, \infty)\), \(f'(x) > 0\) indicating \(f\) is increasing.

Understanding this helps in figuring out the general shape of the function and identifying critical points.

Critical Points

Critical points of a function are where its derivative is zero or undefined. These points are potential locations of local maxima, minima, or points of inflection.
For the function \(f\), \(x = 1\) is a critical point because that's where the behavior changes. At this point, \(f(x)\) stops decreasing and starts increasing. If \(f(1) = 1\), then it forms a local minimum assuming \(f'(1) = 0\), which is characteristic of a smooth transition between decreasing and increasing behavior.
In general, determining critical points is useful when seeking to optimize a function or understand its peak behavior.

Increasing and Decreasing Functions

How a function increases or decreases gives us insights into its overall shape and behavior. When a function is increasing, its output grows as the input grows. Conversely, when it is decreasing, its output falls as the input grows.
According to the exercise, the intervals of increase and decrease help us conclude the behavior of \(f(x)\):

• Decreasing on \((-\infty, 1)\): as \(x\) approaches 1 from the left, \(f(x)\) decreases towards 1.
• Increasing on \((1, \infty)\): as \(x\) moves past 1, \(f(x)\) increases away from 1.

This analysis shows that \(f(x)\) achieves a minimum at \(x = 1\), ensuring that \(f(x) \geq 1\) for all \(x\) based on this transition in behavior.