Problem 34 - Calculus Volume 1

Question

Each pair of functions, find a. \(f+g\) b. \(f-g\) c. \(f \cdot g\) d. \(f / g .\) Determine the domain of each of these new functions. \(f(x)=x-8, g(x)=5 x^{2}\)

Step-by-Step Solution

Answer

a) \(f+g = 5x^2 + x - 8\), \(\text{domain: } \mathbb{R}\) b) \(f-g = -5x^2 + x - 8\), \(\text{domain: } \mathbb{R}\) c) \(f \cdot g = 5x^3 - 40x^2\), \(\text{domain: } \mathbb{R}\) d) \(f/g = \frac{x-8}{5x^2}\), \(\text{domain: } \mathbb{R} \setminus \{0\}\)

Step 1: Find \(f+g\)

To find \(f+g\), simply add \(f(x)\) and \(g(x)\): \(f(x) = x - 8\) and \(g(x) = 5x^2\). So, \(f+g = (x - 8) + (5x^2) = 5x^2 + x - 8\). The domain of both functions \(f(x)\) and \(g(x)\) is all real numbers \(\mathbb{R}\), hence the domain of \(f+g\) is also \(\mathbb{R}\).

Step 2: Find \(f-g\)

To find \(f-g\), subtract \(g(x)\) from \(f(x)\): \(f(x) = x - 8\) and \(g(x) = 5x^2\). So, \(f-g = (x - 8) - (5x^2) = -5x^2 + x - 8\). Similar to the addition, the domain of \(f-g\) is all real numbers \(\mathbb{R}\).

Step 3: Find \(f \cdot g\)

To find \(f \cdot g\), multiply \(f(x)\) and \(g(x)\): \(f(x) = x - 8\) and \(g(x) = 5x^2\). So, \(f \cdot g = (x - 8) \cdot (5x^2) = 5x^2 \cdot x - 5x^2 \cdot 8 = 5x^3 - 40x^2\). The domain remains all real numbers \(\mathbb{R}\) because there's no restriction from either function.

Step 4: Find \(f / g\)

To find \(f / g\), divide \(f(x)\) by \(g(x)\): \(f(x) = x - 8\) and \(g(x) = 5x^2\). So, \(f / g = \frac{x - 8}{5x^2}\). The domain of this function is all real numbers except where \(g(x) = 0\), i.e., \(5x^2 = 0\). Solving for \(x\), we get \(x = 0\). Thus, the domain of \(f/g\) is all real numbers \(\mathbb{R}\) except \(x = 0\).

Key Concepts

Domain of a Function

In mathematics, the domain of a function is the set of all possible input values (usually represented by the variable \(x\)) that the function can accept without causing any issues. These could include any number that doesn't result in a division by zero or a square root of a negative number, for example. When we have functions like \(f(x) = x - 8\) and \(g(x) = 5x^2\), both are polynomial functions.

• Polynomial functions are naturally defined for all real numbers, which means their domain is \(\mathbb{R}\).
• For operations on these functions, it is essential to consider the domain of the resulting function as well.
• For instance, when dividing two functions, you need to ensure the divisor is not zero.

Sum of Functions

Adding functions means that for any input \( x \), we find the sum of their outputs. To compute \(f+g\), you add \(f(x)\) and \(g(x)\). In this exercise:

• \(f(x) = x - 8\) and \(g(x) = 5x^2\).
• The sum is \(f+g = 5x^2 + x - 8\).
• Since both \(f(x)\) and \(g(x)\) are defined for all real numbers, \(f+g\) inherits this domain.

Difference of Functions

The difference of two functions \(f-g\) is found by subtracting the second function from the first for any value of \(x\). Here:

• Given \(f(x) = x - 8\) and \(g(x) = 5x^2\).
• The difference is \(f-g = -5x^2 + x - 8\).
• The domain remains \(\mathbb{R}\) as subtraction doesn't introduce any new restrictions.

Product of Functions

When multiplying functions, each input \(x\) produces a product of their outputs. To find \(f \cdot g\):

• Multiply \(f(x) = x - 8\) by \(g(x) = 5x^2\).
• The result is: \(f \cdot g = 5x^3 - 40x^2\).
• As with sums and differences, the domain remains all real numbers, \(\mathbb{R}\), because no new restrictions are introduced by the operation.

Quotient of Functions

Dividing functions involves dividing their outputs for each input \(x\). For \(f/g\), it becomes essential to ensure the denominator is non-zero:

• \(f(x) = x - 8\) and \(g(x) = 5x^2\) become \(f/g = \frac{x - 8}{5x^2}\).
• The domain is all real numbers except where \(g(x) = 0\).
• Solving \(5x^2 = 0\) reveals \(x = 0\), which must be excluded.

Therefore, the domain of \(f/g\) is \(\mathbb{R}\ \backslash \{0\}\). Excluding these points ensures the function remains valid.