Question

A function \(f\) has a simple zero (or a zero of multiplicity 1) at \(x_{0}\) if \(f\) is differentiable in a neighborhood of \(x_{0}\) and \(f\left(x_{0}\right)=0,\) while \(f^{\prime}\left(x_{0}\right) \neq 0\). (a) Prove that \(f\) has a simple zero at \(x_{0}\) if and only if $$ f(x)=g(x)\left(x-x_{0}\right) $$ where \(g\) is continuous at \(x_{0}\) and differentiable on a deleted neighborhood of \(x_{0}\), and \(g\left(x_{0}\right) \neq 0 .\) (b) Give an example showing that \(g\) in(a) need not be differentiable at \(x_{0}\) -

Short answer

Based on the given step by step solution, provide a short answer: The function \(f(x)\) has a simple zero at \(x_0\) if and only if \(f(x)=g(x)(x-x_0)\) with \(g(x)\) continuous at \(x_0\), differentiable on a deleted neighborhood of \(x_0\), and \(g(x_0)\neq 0\). For example, the function \(f(x) = (x^2 - 1)|x - 1|\) has a simple zero at \(x_0 = 1\) with the corresponding function \(g(x) = (x + 1)|x - 1|\) which is continuous at \(x_0 = 1\) but not differentiable at \(x_0 = 1\), demonstrating that \(g(x)\) need not be differentiable at \(x_0\).

Step-by-step solution

Part (a): Proof of the condition for a simple zero

To show that \(f(x)\) has a simple zero at \(x_0\) if and only if \(f(x) = g(x)(x-x_{0})\), where \(g(x)\) is continuous at \(x_0\) and differentiable on a deleted neighborhood of \(x_0\), we need to prove both directions: 1. If \(f(x)\) has a simple zero at \(x_0\), then \(f(x) = g(x)(x-x_{0})\). 2. If \(f(x) = g(x)(x-x_{0})\), where \(g(x)\) is continuous at \(x_0\) and differentiable on a deleted neighborhood of \(x_0\), then \(f(x)\) has a simple zero at \(x_0\). Proof for 1: Since \(f(x)\) has a simple zero at \(x_0\), we know that \(f(x_0)=0\) and \(f'(x_0)\neq0\). Let's define the function \(g(x)=\frac{f(x)}{x-x_0}\) for \(x\neq x_0\). Since \(f(x_0)=0\), we can define \(g(x_0)=\lim_{x\to x_0} g(x)\). Using L'Hôpital's rule to compute the limit, we get: $$ g(x_0) = \lim_{x\to x_0} \frac{f(x)}{x-x_0} = \lim_{x\to x_0} \frac{f'(x)}{1} = f'(x_0) \neq 0. $$ Thus, \(g(x)\) is continuous at \(x_0\). Since \(f(x)\) is differentiable in a neighborhood of \(x_0\), and the derivate of \(x - x_0\) is 1, then by the quotient rule, \(g(x)\) is differentiable on a deleted neighborhood of \(x_0\). Hence, we have shown that \(f(x) = g(x)(x-x_{0})\). Proof for 2: Now, let's assume that \(f(x)=g(x)(x-x_{0})\), where \(g(x)\) is continuous at \(x_0\) and differentiable on a deleted neighborhood of \(x_0\). Then, \(f(x_0) = g(x_0)(x_0-x_0) = 0\). Using the product rule, we compute the derivative of \(f(x)\): $$ f'(x) = g'(x)(x-x_{0}) + g(x). $$ Since \(g(x_0)\neq 0\) and \(g'(x)\) is differentiable at \(x_0\), then \(f'(x_0) = g'(x_0)(0) + g(x_0) = g(x_0) \neq 0\). Therefore, \(f(x)\) has a simple zero at \(x_0\).

Part (b): Example where g(x) is not differentiable at \(x_0\)

Now, let's provide an example of a function \(f(x)\) that satisfies the required condition but \(g(x)\) is not differentiable at \(x_0\). Let \(f(x) = x^2(x^2 - 1)\) and \(x_0 = 1\). Then, \(f(1) = 0\). We define \(g(x) = x^2\) for \(x \neq 1\). We can see that \(g(1) = 1 \neq 0\), and that \(g(x)\) is differentiable on a deleted neighborhood of \(1\). However, \(g(x) = x^2\) is differentiable at \(x_0=1\). So, let's pick a less trivial example. Let \(f(x) = (x^2 - 1)|x - 1|\) and \(x_0 = 1\). Then, \(f(x) = (x - 1)(x + 1)|x - 1|\). Define \(g(x) = (x + 1)|x - 1|\) for \(x \neq 1\). Again, we can see that \(g(1) = 2 \neq 0\). However, \(g(x)\) is not differentiable at \(x_0=1\) due to the absolute value function. This example shows that \(g(x)\) need not be differentiable at \(x_0\).

Key concepts

Multiplicity of Zeros

Understanding the multiplicity of zeros is fundamental when studying the behavior of functions near their roots. When we say a function f has a simple zero at a point x₀, we mean that the function crosses the x-axis at this point and does not just touch it and turn around. The term multiplicity refers to the number of times a zero is repeated for a function at a given point.

A zero of multiplicity 1, or a simple zero, has some important characteristics. Firstly, the function value at this point is zero, i.e., f(x₀) = 0. Additionally, the derivative of the function at this point is non-zero, f'(x₀) ≠ 0, suggesting the function is sloping away from the x-axis instead of just tangentially touching it.

Differentiability of Functions

The property of differentiability is closely tied to the behavior of functions and their graphs. For a function to be differentiable at a point x₀, it must be smooth at that point with no sharp corners or cusps. Differentiability implies continuity, but not vice versa. This means that a differentiable function at x₀ is automatically continuous at that point; however, if a function is continuous at x₀, it is not necessarily differentiable there.

Differentiability and Simple Zero

When exploring the context of a simple zero, the concept of differentiability becomes crucial since we need the slope, f'(x₀), to be non-zero. Moreover, using the derivative information, we can determine whether f is increasing or decreasing at x₀, thus confirming the presence of a simple zero.

Continuity of Functions

Continuity is a key concept in calculus and describes a function that is unbroken or uninterrupted. For a function f to be continuous at a point x₀, three conditions must be met: f(x₀) is defined, the limit as x approaches x₀ of f(x) exists, and both are equal. In simple terms, you can draw the function at this point without lifting your pencil off the paper.

Continuity is vital when dealing with a simple zero. The accompanying function g, in the form f(x) = g(x)(x-x₀), must be continuous at x₀ so as not to disrupt the smooth transition through the zero at x₀.

L'Hôpital's Rule

Often when calculating limits, we encounter indeterminate forms like 0/0 or ∞/∞. L'Hôpital's rule is a tool that helps us resolve these indeterminate forms by comparing the rates of change of the numerator and denominator functions. The rule states that if the limits lim_x→cf(x) and lim_x→cg(x) both approach 0 or both approach ∞, then lim_x→c(f(x)/g(x)) = lim_x→c(f'(x)/g'(x)) under specific conditions where f and g are differentiable near c and g'(x) does not equal zero near c.

Understanding L'Hôpital's rule is advantageous when proving a function has a simple zero, as shown in the exercise. It allows us to define g(x₀) at the zero point for f(x) by examining the limit and the behaviors of both f(x) and its derivative f'(x) around that point.

Product Rule for Differentiation

The product rule is an essential differentiation rule that states when you have a product of two functions, say u(x) and v(x), the derivative of their product is given by u'(x)v(x) + u(x)v'(x). It is crucial for finding derivatives of functions that are multiplied together, which is especially common when dealing with zeros of multiplicity higher than one.

In the context of our exercise, we use the product rule to differentiate the function f(x) = g(x)(x-x₀). Applying the product rule allows us to establish the conditions under which f(x) has a simple zero at x₀ by ensuring that the derivative at that point is non-zero, which is exactly what we need to prove such that