Problem 53 - Calculus and its applications

Question

Let \(f(x)=\frac{x^{2}+4 x+3}{x+1}\). A student recognizes that this function can be simplified as $$f(x)=\frac{x^{2}+4 x+3}{x+1}=\frac{(x+1)(x+3)}{x+1}=x+3$$ Since \(y=x+3\) is a line with slope 1 , the student makes the following conclusions: \(f^{\prime}(-2)=1\), \(f^{\prime}(-1)=1, f^{\prime}(0)=1, f^{\prime}(1)=1 .\) Where did the student make an error?

Step-by-Step Solution

Answer

The student incorrectly claims \(f'(-1)=1\); \(f\) is undefined at \(x=-1\), so the derivative does not exist there.

Step 1: Understanding the Simplification

The function \(f(x)=\frac{x^2+4x+3}{x+1}\) can be simplified to \(f(x) = \frac{(x+1)(x+3)}{x+1} = x+3\). However, this simplification holds true only when \(x eq -1\), because the denominator should not be zero. Hence, \(f(x)\) is undefined at \(x = -1\).

Step 2: Recognizing Discontinuity

The simplification \(f(x) = x+3\) suggests that the function is equivalent to the line \(y = x + 3\) everywhere except \(x = -1\). Therefore, at \(x=-1\), \(f(x)\) has a point of discontinuity, meaning the function does not have a derivative (slope) at this point.

Step 3: Calculating the Derivative

For \(x eq -1\), the derivative \(f'(x)\) of the simplified function \(f(x) = x + 3\) is simply \(1\), since the derivative of \(x+3\) with respect to \(x\) is \(1\). However, at \(x = -1\), the derivative does not exist because the function is undefined there.

Step 4: Identifying Errors

The student's error lies in concluding that \(f'(-1) = 1\). While \(f'(x) = 1\) holds for \(x eq -1\), it does not exist at \(x = -1\) due to the discontinuity (division by zero in original fraction). The evaluations of \(f'(-2)\), \(f'(0)\), and \(f'(1)\) are correct, as they do not correspond to the point of discontinuity.

Key Concepts

Discontinuity in Functions

When we examine functions in calculus, particularly rational functions, a point of discontinuity occurs when the function is undefined. This typically happens when the denominator of a rational function equals zero. In the given function, \(f(x) = \frac{x^2 + 4x + 3}{x + 1}\), simplifying this function results in \(f(x) = x + 3\) for all \(x\) except \(x = -1\). Why not \(x = -1\)? Notice that substituting \(x = -1\) into the original function would cause division by zero, making it undefined at this point.

Thus, the function \(f(x)\) has a discontinuity at \(x = -1\). Such points are crucial because they indicate that you cannot calculate a derivative there. It's like the function "jumps" over this value on the graph. These breakpoints might not be apparent in the simplified form, so it’s essential to check the original function to identify any discontinuities.

Derivative at a Point

In calculus, the derivative at a point describes the rate at which a function changes as the input changes. For most values, this rate of change or slope can be determined using standard differentiation rules. However, the derivative doesn’t exist at points of discontinuity.

For example, with \(f(x) = \frac{(x+1)(x+3)}{x+1}\), when simplified to \(x + 3\), it suggests a constant slope. The derivative of \(x + 3\) is 1. Hence, the slope or rate of change is 1 for all \(xeq-1\). The point-at-discontinuity--at \(x=-1\) where the denominator initially zeroes out--the derivative isn’t defined. It's like looking at a smooth ramp that suddenly drops with no gradual slope change; the derivative "disappears" at these jumps.

Therefore, a common mistake is assuming that simplifications reflect the true derivative at all points. Rather than blindly applying rules, examine potential discontinuities first.

Rational Functions Simplification

Rational functions, which are ratios of two polynomials, can sometimes be simplified by canceling common factors from the numerator and the denominator. In the given exercise, \(f(x) = \frac{x^2 + 4x + 3}{x + 1}\) simplifies to \(x + 3\) by factoring the numerator's quadratic expression into \((x + 1)(x + 3)\) and canceling \((x + 1)\) from the denominator.

You might think this process is perfectly clean, but it can lead to misconceptions if not careful. What seems like a simple line graph with a steady slope ignores potential undefined points. To avoid errors when simplifying rational functions, always:

• Check conditions where the denominator equals zero and recognize them as potential discontinuities.
• Factor completely before canceling to ensure every step resolves actual values.
• Understand that the simplification can mask points where the function is undefined, leading to errors if applying derivatives without full consideration.

By observing these methods, you can more accurately work with rational functions, understanding both their continuous parts and discontinuities.