Problem 58 - University Calculus: Early Transcendentals

Question

Is there a value of \(b\) that will make $$g(x)=\left\\{\begin{array}{ll} x+b, & x<0 \\ \cos x, & x \geq 0 \end{array}\right.$$ continuous at \(x=0 ?\) Differentiable at \(x=0 ?\) Give reasons for your answers.

Step-by-Step Solution

Answer

For \(b = 1\), \(g(x)\) is continuous at \(x = 0\), but not differentiable.

Step 1: Understand the Problem

To determine if the function \(g(x)\) is continuous at \(x = 0\), we need to ensure that the limit as \(x\) approaches 0 from the left equals the function value at 0. For differentiability, the left and right derivatives at \(x = 0\) must be equal.

Step 2: Check Continuity at x=0

A function is continuous at a point if the limit from the left equals the limit from the right and both equal the function value there. For continuity at \(x = 0\):- Left-hand limit: \(\lim_{x \to 0^-} (x + b) = b\).- Right-hand limit: \(\lim_{x \to 0^+} \cos x = \cos 0 = 1\).The function value at \(x = 0\) is \(\cos 0 = 1\).For continuity, set the left-hand limit equal to the function value: \(b = 1\).

Step 3: Check Differentiability at x=0

A function is differentiable at a point if the left and right derivatives exist and are equal at that point.- Left-hand derivative at \(x = 0\): The derivative of \(x + b\) is \(\lim_{h \to 0^-} \frac{(h + b) - b}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1\).- Right-hand derivative at \(x = 0\): The derivative of \(\cos x\) is \(-\sin x\), so \(\lim_{h \to 0^+} \frac{\cos h - 1}{h} = 0\). Since \(1 eq 0\), \(g(x)\) is not differentiable at \(x=0\).

Key Concepts

Differentiability

Differentiability refers to whether a function has a derivative at a specific point. A function is differentiable at a point if it has a smooth graph without any sharp corners or cusps at that point.
To check for differentiability at any point, both the derivative from the left and the derivative from the right must exist and be equal. For the function \( g(x) \) given in the exercise, we found the left and right derivatives at \( x = 0 \).
The left derivative from \( x < 0 \) is 1, as derived from the linear expression \( x + b \), which simplifies to \( \lim_{h \to 0^-} \frac{h}{h} = 1 \). The right derivative from \( \cos x \) at \( x \geq 0 \) leads to a limit of 0, \( \lim_{h \to 0^+} \frac{\cos h - 1}{h} = 0 \).
Since the left and right derivatives are not the same, \( g(x) \) is not differentiable at \( x = 0 \). Differentiability goes hand-in-hand with having a single tangent line at the point, which is impossible here due to the difference in slope from the left and right.

Limits

The concept of limits helps us understand the behavior of functions as they approach specific points. In this exercise, we explore the limit of \( g(x) \) as \( x \to 0 \) from both directions

• The left-hand limit is the value \( g(x) \) approaches as \( x \) approaches 0 from the negative side. For \( x < 0 \), the function is \( x + b \), so \( \lim_{x \to 0^-} (x + b) = b \).
• The right-hand limit for \( x \geq 0 \) is \( \cos x \), which simplifies to \( \lim_{x \to 0^+} \cos x = \cos 0 = 1 \).

This illustrates that when discussing limits at a point, it is important that the approach from either side achieves the same result for continuity. In the case of differentiability, although not directly using limits, the process heavily involves them as evident in the derivative calculations.

Function Behavior

The way a function behaves around a point is critical in understanding its continuity and differentiability. A function can change its definition as it crosses into different domains.
For \( g(x) \), we have distinct behavior depending if \( x \) is less than or greater than or equal to zero:

• When \( x < 0 \), \( g(x) = x + b \), which is a linear function.
• When \( x \geq 0 \), \( g(x) = \cos x \), which is periodic and smooth.

These two pieces highlight an important aspect: a function's behavior tells us where potential discontinuities and non-differentiable points occur. Although the linear and trigonometric parts are individually consistent within their intervals, their differing slopes at 0 complicate differentiability there.

Derivatives

Derivatives measure a function's rate of change or the slope of its graph at a particular point. A connected question to both continuity and differentiability is whether you can calculate this slope effectively.
In the exercise, determining the derivatives involved finding limits of functions as \( h \) approaches 0 for both sides of the point.

• The calculation for \( x < 0 \) gives a derivative of 1 by simplifying \( \lim_{h \to 0^-} \frac{h}{h} \), indicating a constant slope.
• For \( x \geq 0 \), \( \cos x \) has a derivative involving \( -\sin x \) as \( \lim_{h \to 0^+} \frac{\cos h - 1}{h} \), which approaches 0.

The mismatch in derivatives suggests a difference in rate of change, reflecting varied behavior of the function parts. These principles underline that when derivatives vary on either side of a point, a function cannot be smooth or tangent consistently across that divide.