Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

The given function is not defined at a certain point. How should it be defined in order to make it continuous at that point? $$ \phi(x)=\frac{x^{4}+2 x^{2}-3}{x+1} $$

Step-by-Step Solution

Answer

Define \( \phi(-1) = -2 \) to make the function continuous at \( x = -1 \).

Step 1: Identify the Point of Discontinuity

The function \( \phi(x) = \frac{x^4 + 2x^2 - 3}{x+1} \) is undefined at \( x = -1 \), since the denominator becomes zero at this point.

Step 2: Factor the Numerator

To find if there is a removable discontinuity at \( x = -1 \), factor the numerator \( x^4 + 2x^2 - 3 \). This may involve using polynomial division or recognizing a pattern. For simplicity, check if it is factorable by simpler components first.

Step 3: Polynomial Division

Perform polynomial division of the numerator \( x^4 + 2x^2 - 3 \) by \( x+1 \). This helps find whether (x + 1) is a factor. By dividing, the quotient is \( x^3 - x^2 + x - 1 \), and the remainder is 0, confirming that \( (x+1) \) is a factor.

Step 4: Rewrite the Function

Since \( (x+1) \) is a factor, rewrite \( \phi(x) \) as \( \phi(x) = x^3 - x^2 + x - 1 \) for \( x eq -1 \), removing the discontinuity.

Step 5: Define Function at \( x = -1 \)

To make the function continuous at \( x = -1 \), we define \( \phi(-1) \) as the limit of \( \phi(x) \) as \( x \to -1 \). Calculate \( \phi(-1) = (-1)^3 - (-1)^2 + (-1) - 1 = -1 + 1 - 1 - 1 = -2 \).

Step 6: Final Definition

Thus, to make the function continuous at \( x = -1 \), define \( \phi(-1) = -2 \). The function can be written as \( \phi(x) = x^3 - x^2 + x - 1 \) \( \forall x \).

Key Concepts

Point of Discontinuity

A point of discontinuity in a function occurs where the function is not defined, or where there is an abrupt change in its value. In the function \( \phi(x) = \frac{x^4 + 2x^2 - 3}{x+1} \), the point of discontinuity is at \( x = -1 \). This is because the denominator \( x+1 \) equals zero at \( x = -1 \), making the function undefined at this point. When the denominator of a fraction is zero, it causes the function to be discontinuous at that value of \( x \). Almost like a gap that needs to be filled for the function to remain smooth and continuous.

Polynomial Division

Polynomial division is a method used to divide one polynomial by another, much like long division with numbers. In our exercise, we need to determine if the polynomial in the numerator \( x^4 + 2x^2 - 3 \) can be divided by \( x+1 \). This helps check if \( x+1 \) is a factor of the polynomial.
We divide \( x^4 + 2x^2 - 3 \) by \( x+1 \) and find the quotient \( x^3 - x^2 + x - 1 \) with a remainder of 0. A remainder of 0 means \( x+1 \) is a factor, which is crucial information. It indicates that the point of discontinuity at \( x = -1 \) may be removable, as the numerator and the denominator have a common factor.

Removable Discontinuity

A removable discontinuity happens when a function has a hole at a certain point, meaning the function value can be defined to fill this gap smoothly. In the function \( \phi(x) \), this discontinuity at \( x = -1 \) can be removed because \( x+1 \) is a factor of both the numerator and the denominator. This allows us to cancel out \( x+1 \) from both the top and the bottom, thereby 'removing' the discontinuity.
After cancellation, the function becomes \( x^3 - x^2 + x - 1 \), valid for all \( x eq -1 \). We can then define the function value at \( x = -1 \) such that it is continuous.

Limit of a Function

The limit of a function, as \( x \) approaches a certain value, gives us the behavior of the function near that point. It is especially useful when addressing discontinuities. In our scenario, to make the function \( \phi(x) \) continuous at \( x = -1 \), we use the limit to determine what the function value should be at this point to remove the discontinuity.
We calculate the limit \( \lim\limits_{x \to -1} (x^3 - x^2 + x - 1) \). Substituting \( x = -1 \) into the simplified function, the result is \( -2 \). Therefore, defining \( \phi(-1) = -2 \) ensures that the previously discontinuous point \( x = -1 \) is now smoothly connected in the graph of the function.