Problem 25 - Calculus

Question

give an example of: A rational function that has zeros at \(x=\pm 1\) and is not differentiable at \(x=\pm 2\)

Step-by-Step Solution

Answer

\( f(x) = \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)} \) satisfies the conditions.

Step 1: Define the Zeros

To create a rational function with zeros at \( x = \pm 1 \), we start by including \((x - 1)\) and \((x + 1)\) in the numerator of our function. This will ensure that when \(x = 1\) or \(x = -1\), the numerator equals zero, giving us the desired zeros.

Step 2: Ensure Non-differentiability at Specific Points

To ensure the function is not differentiable at \( x = \pm 2 \), we need the denominator to become zero at these points. This can be achieved by including the factors \((x - 2)\) and \((x + 2)\) in the denominator. The function will have vertical asymptotes at these points, making it non-differentiable.

Step 3: Construct the Rational Function

Using the factors outlined in Steps 1 and 2, we construct the rational function as follows:\[ f(x) = \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)} \] This function satisfies all the conditions: it has zeros at \(x = 1\) and \(x = -1\), and is non-differentiable at \(x = 2\) and \(x = -2\).

Key Concepts

Zeros of a function

In the world of mathematics, understanding the zeros of a function is crucial. Zeros, also known as roots, are the points where the function equals zero. For a function \(f(x)\), if \(f(a) = 0\), then \(x = a\) is a zero of the function. In the context of our rational function \(f(x) = \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)}\), the zeros are at the points where the numerator becomes zero.

Let's break it down further:

• The numerator \((x - 1)(x + 1)\) equals zero when \(x = 1\) or \(x = -1\).
• This is because multiplying zero by any number results in zero, which is why these values are the zeros.

When you set \(f(x)\) equal to zero, the denominator does not play a role in determining the zeros of the function. It only affects where the function is undefined. Understanding zeros helps us in graphing the function and predicting its behavior around these points.

Differentiability

Differentiability is a key concept in calculus that tells us if a function has a derivative at a certain point. If a function is differentiable at a point, it means the function's graph has a tangent at that point, and we can find an instantaneous rate of change.

However, our function \(f(x) = \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)}\) is designed to be non-differentiable at \(x = 2\) and \(x = -2\). But why?

• The reason lies in the denominator: \((x - 2)(x + 2)\).
• A function is undefined where its denominator is zero, leading to points of non-differentiability.

Non-differentiable points often lead to sudden "jumps" or "breaks" in the graph. At these points, the slope of the function does not exist. Understanding differentiability helps in analyzing the smoothness and continuity of a graph.

Vertical Asymptotes

Vertical asymptotes are lines that a graph approaches but never actually touches. They occur in rational functions where the denominator is zero, indicating points where the function is undefined. For our rational function \(f(x) = \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)}\), there are vertical asymptotes at \(x = 2\) and \(x = -2\).

Here's why that happens:

• Since division by zero is undefined, the function heads towards infinity as \(x\) approaches \(2\) or \(-2\).
• You will notice that the graph of the function will get closer and closer to the lines \(x = 2\) and \(x = -2\) as \(x\) gets closer to these values, but will never actually reach them.

Vertical asymptotes are essential for identifying the behavior of a function near its undefined points and understanding the graph's overall shape and direction.