Problem 22 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 34, the function is the set of all ordered pairs \((x, y)\) satisfying the given equation. Find the domain and range of the function, and draw a sketch of the graph of the function. $$ f: y=\frac{x^{3}+3 x^{2}+x+3}{x+3} $$

Step-by-Step Solution

Answer

Domain: \( (-\infty, -3) \cup (-3, \infty) \), Range: \( [1, \infty) \)

Step 1: Simplify the Function

First, simplify the given function. \( f(x) = \frac{x^3 + 3x^2 + x + 3}{x + 3} \)Using polynomial division, or recognizing that the numerator can be factored, we find: \( x^3 + 3x^2 + x + 3 = (x+3)(x^2+0x+1) \)So, \( f(x) = \frac{(x+3)(x^2+1)}{x+3} = x^2+1 \) for \( x eq -3 \)

Step 2: Find the Domain

The domain of the function includes all real numbers except where the denominator is zero. Here, the function is undefined at \( x = -3 \). Therefore, the domain is: \( (-\infty, -3) \cup (-3, \infty) \)

Step 3: Find the Range

Since the simplified function \( y = x^2 + 1 \) can produce any value for \( y \) greater than or equal to 1, the range is: \( [1, \infty) \)

Step 4: Sketch the Graph

To sketch the graph, plot the function \( y = x^2 + 1 \) which is a standard parabola shifted 1 unit upwards. Note that there is a hole in the graph at \( x = -3 \).

Key Concepts

Function Simplification

Simplifying a function is a crucial step in understanding its behavior. For the given function:
\( f(x) = \frac{x^{3} + 3x^{2} + x + 3}{x + 3} \), we simplify it by factoring the numerator. We notice that the numerator can be factored as \((x + 3)(x^{2} + 1) \).
This allows us to cancel out the common \(x+3 \) term from the numerator and the denominator:
\[ f(x) = \frac{(x+3)(x^{2}+1)}{x+3} = x^2 + 1 \]
However, this simplification is valid only for \( x eq -3 \). Automatically, we exclude \(x = -3 \) from the function’s domain, since the value would make the denominator zero, causing an undefined value.

Domain of a Function

The domain of a function consists of all input values (\( x \)) for which the function is defined. For our simplified function,
\( f(x) = x^2 + 1 \).
We start with the generalized fact that polynomial functions like \(x^2 + 1 \) are defined for all real numbers.
However, because we simplified the function from an original form with a denominator of \(x+3 \), the domain is restricted by excluding \(x=-3\) where the denominator is zero.
Thus, the domain of the function is all real numbers except \(x=-3\):
\[ (-\infty, -3) \cup (-3, \infty) \] \;.

Range of a Function

The range of a function includes all possible output values (\(y\)) for given input values (\(x\)). For the simplified function:
\(y = x^2 + 1\), we recognize that it is a quadratic function shifted one unit upwards.
Evaluating \(y = x^2 + 1\), we can see that since \(x^2\) is always non-negative, \(x^2 + 1\) will always be at least 1.
Hence, the smallest value \(y\) can take is 1 (which happens when \(x = 0\)).
As \(x\) increases or decreases without bound, \(y\) increases indefinitely.
Thus, the range of the function is:
\[ [1, \infty) \].

Graph Sketching

To visualize the function, we start by sketching the graph of the simplified form \(y = x^2 + 1\).

• This is a standard parabolic graph.
• It opens upwards.
• It is shifted 1 unit up compared to \( y = x^2 \).

The graph passes through the point \((0, 1)\).
Also, remember that since \(x=-3\) was excluded from our domain, there is a hole in the graph where \(x = -3\) would have been. This indicates a discontinuity at that point.
For a complete sketch:
- Draw the standard parabola \(y = x^2 + 1\). - Indicate a hole on this graph at the point where \((x = -3, y = (-3)^2 + 1 = 10)\) because the function is undefined at \(x = -3\).