Problem 4 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 3-8: Find (a) \((f+g)(x),(f-g)(x),(f g)(x)\), and \((f / g)(x)\) (b) the domain of \(f+g, f-g\), and \(f g\) (c) the domain of \(f / g\) $$ f(x)=x^{2}+x, \quad g(x)=x^{2}-3 $$

Step-by-Step Solution

Answer

(f+g)(x)=2x^2+x-3, (f-g)(x)=x+3, (fg)(x)=x^4+x^3-3x^2-3x; domain for f+g, f-g, fg: all real numbers; domain for f/g: all real numbers except ±√3.

Step 1: Calculate \((f+g)(x)\)

The expression \((f + g)(x)\) is calculated by adding the functions \(f(x)\) and \(g(x)\):\[(f+g)(x) = f(x) + g(x) = (x^2 + x) + (x^2 - 3)\]Combine like terms:\[(f+g)(x) = 2x^2 + x - 3\]

Step 2: Calculate \((f-g)(x)\)

The expression \((f - g)(x)\) is calculated by subtracting the function \(g(x)\) from \(f(x)\):\[(f-g)(x) = f(x) - g(x) = (x^2 + x) - (x^2 - 3)\]Combine like terms:\[(f-g)(x) = x + 3\]

Step 3: Calculate \((fg)(x)\)

The expression \((fg)(x)\) is calculated by multiplying the functions \(f(x)\) and \(g(x)\):\[(fg)(x) = f(x) \times g(x) = (x^2 + x)(x^2 - 3)\]Expand the product:\[(fg)(x) = x^4 - 3x^2 + x^3 - 3x\]Reorder the terms:\[(fg)(x) = x^4 + x^3 - 3x^2 - 3x\]

Step 4: Calculate \((f/g)(x)\)

The expression \((f/g)(x)\) is calculated by dividing the function \(f(x)\) by \(g(x)\):\[(f/g)(x) = \frac{f(x)}{g(x)} = \frac{x^2 + x}{x^2 - 3}\]

Step 5: Determine the domain of \(f+g\), \(f-g\), and \(fg\)

The domain of the functions \(f+g\), \(f-g\), and \(fg\) are determined by finding values of \(x\) where both \(f(x)\) and \(g(x)\) are defined. Both \(f(x) = x^2 + x\) and \(g(x) = x^2 - 3\) are polynomials, which are defined for all real numbers. Thus, the domain is:\[(-\infty, \infty)\]

Step 6: Determine the domain of \(f/g\)

The domain of the function \(f/g\) is determined by finding the values of \(x\) for which the denominator \(g(x)\) is not zero. We solve:\[g(x) = x^2 - 3 = 0\]\[x^2 = 3\]\[x = \pm \sqrt{3}\]Thus, the domain of \(f/g\) is all real numbers except \(x = \sqrt{3}\) and \(x = -\sqrt{3}\):\[(-\infty, -\sqrt{3}) \cup (-\sqrt{3}, \sqrt{3}) \cup (\sqrt{3}, \infty)\]

Key Concepts

Domain of Function

The domain of a function refers to all the possible input values (usually represented as \(x\)) that will not cause the function to be undefined.
For polynomial functions like the ones given in this exercise, \(f(x) = x^2 + x\) and \(g(x) = x^2 - 3\), the expressions are defined for all real numbers since they do not involve division by zero or negative square roots. However, when dealing with operations involving these functions such as division, we must be careful. For \(\frac{f}{g}(x)\), the values where \(g(x)\) equals zero must be excluded from the domain.
Therefore, we solve \(x^2 - 3 = 0\) to find that \(x = \pm \,\sqrt{3}\) are points that make \(g(x)\) zero.
Thus, the domain of \(f/g\) is all real numbers except \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).

Polynomial Functions

Polynomial functions are expressions consisting of variables and coefficients, with operations of addition, subtraction, multiplication, and non-negative integer exponents.
They have forms such as \(x^2 + x\) and \(x^2 - 3\), as seen in the current exercise. Polynomial functions are smooth, continuous curves spanning all real numbers.
They do not have vertical asymptotes or breaks meaning they are defined for any value of \(x\). In operations, whether adding, subtracting, multiplying, or dividing polynomials, the outcome is usually another polynomial, as long as division doesn’t involve dividing by zero.

Addition of Functions

Adding two functions involves summing the expressions of both functions, which means combining their like terms. For the functions \(f(x) = x^2 + x\) and \(g(x) = x^2 - 3\), we perform:

• Add the \(x^2\) terms together to get \(2x^2\).
• Keep the \(x\) term from \(f(x)\), as there is no corresponding term in \(g(x)\).
• Combine or adjust any constant terms.

Thus, \((f+g)(x) = 2x^2 + x - 3\).
This new function is also a polynomial and inherits the domain of being defined for all real numbers.

Subtraction of Functions

Subtracting one function from another involves taking the second function’s terms and essentially adding their negatives to the first.
This means combining like terms, but with the opposite sign for those from the subtracted function. When subtracting \(g(x) = x^2 - 3\) from \(f(x) = x^2 + x\):

• The \(x^2\) terms cancel each other out.
• The \(-x^2\) from \(g(x)\) turns into subtraction of \(x^2\), effectively canceling.
• The \(x\) remains from \(f(x)\).
• Add 3 from the constants, because the constant in \(g(x)\) is subtracted as a negative.

This results in \((f-g)(x) = x + 3\).
Again, this function remains a polynomial and is defined for all real numbers.

Multiplication of Functions

Multiplying functions involves distributing each term of one function to every term of the other.
This process is known as the distributive property. For \(f(x) = x^2 + x\) and \(g(x) = x^2 - 3\), the multiplication is:

• \(x^2 \times x^2 = x^4\).
• \(x^2 \times -3 = -3x^2\).
• Additional terms \(x \times x^2 = x^3\) and \(x \times -3 = -3x\).

Combine the resulting terms giving the new polynomial: \((fg)(x) = x^4 + x^3 - 3x^2 - 3x\).
This multiplied function remains a polynomial, with the domain being all real numbers.

Division of Functions

The division of functions is a bit more tricky, primarily because it can lead to undefined values when the denominator is zero. For dividing \(f(x)\) by \(g(x)\): \(\frac{f(x)}{g(x)} = \frac{x^2 + x}{x^2 - 3}\), we must identify the values of \(x\) for which the denominator becomes zero.

• Solve the equation \(x^2 - 3 = 0\) which gives \(x = \pm \sqrt{3}\).

These \(x\) values need to be excluded from the domain since division by zero is undefined. Thus, the domain of \(f/g\) is all real numbers except \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).
This concept helps prevent erroneous outcomes in functional operations.