Problem 5 - Thomas Calculus

Question

If \(f(x)=x+5\) and \(g(x)=x^{2}-3,\) find the following. $$ \begin{array}{ll}{\text { a. } f(g(0))} & {\text { b. } g(f(0))} \\ {\text { c. } f(g(x))} & {\text { d. } g(f(x))} \\ {\text { e. } f(f(-5))} & {\text { f. } g(g(2))} \\ {\text { g. } f(f(x))} & {\text { h. } g(g(x))}\end{array} $$

Step-by-Step Solution

Answer

a. 2, b. 22, c. \(x^2 + 2\), d. \(x^2 + 10x + 22\), e. 5, f. -2, g. \(x + 10\), h. \(x^4 - 6x^2 + 6\).

Step 1: Evaluate g(0)

To find \( f(g(0)) \), we first need to evaluate \( g(0) \). The function \( g(x) = x^2 - 3 \). Substitute \( x = 0 \) to get \( g(0) = 0^2 - 3 = -3 \).

Step 2: Calculate f(g(0))

Now, we need to find \( f(g(0)) \), where \( g(0) = -3 \). Since \( f(x) = x + 5 \), substitute \( x = -3 \) to get \( f(-3) = -3 + 5 = 2 \).

Step 3: Evaluate f(0)

For \( g(f(0)) \), we first calculate \( f(0) \). Using \( f(x) = x + 5 \), substitute \( x = 0 \) to get \( f(0) = 0 + 5 = 5 \).

Step 4: Calculate g(f(0))

Now find \( g(f(0)) \), and we have \( f(0) = 5 \). Substitute into \( g(x) = x^2 - 3 \) to get \( g(5) = 5^2 - 3 = 25 - 3 = 22 \).

Step 5: Calculate f(g(x))

Substitute \( g(x) = x^2 - 3 \) into \( f \), giving \( f(g(x)) = f(x^2 - 3) \). Using \( f(x) = x + 5 \), we get \( f(x^2 - 3) = (x^2 - 3) + 5 = x^2 + 2 \).

Step 6: Calculate g(f(x))

Substitute \( f(x) = x + 5 \) into \( g \), giving \( g(f(x)) = g(x + 5) \). Using \( g(x) = x^2 - 3 \), it becomes \( (x + 5)^2 - 3 \). Expand to get \( x^2 + 10x + 25 - 3 = x^2 + 10x + 22 \).

Step 7: Evaluate f(-5)

For \( f(f(-5)) \), first calculate \( f(-5) \). Use \( f(x) = x + 5 \) and substitute \( x = -5 \) to get \( f(-5) = -5 + 5 = 0 \).

Step 8: Calculate f(f(-5))

Now find \( f(f(-5)) \), knowing \( f(-5) = 0 \). Substitute into \( f(x) = x + 5 \) to get \( f(0) = 0 + 5 = 5 \).

Step 9: Evaluate g(2)

Calculate \( g(g(2)) \) starting with \( g(2) \). With \( g(x) = x^2 - 3 \), substitute \( x = 2 \) to find \( g(2) = 2^2 - 3 = 4 - 3 = 1 \).

Step 10: Calculate g(g(2))

Find \( g(g(2)) \), where \( g(2) = 1 \). Substitute into \( g(x) = x^2 - 3 \) giving \( g(1) = 1^2 - 3 = 1 - 3 = -2 \).

Step 11: Calculate f(f(x))

Substitute \( f(x) = x + 5 \) into itself to get \( f(f(x)) = f(x + 5) \). From \( f(x) = x + 5 \), it becomes \( (x + 5) + 5 = x + 10 \).

Step 12: Calculate g(g(x))

Substitute \( g(x) = x^2 - 3 \) into itself for \( g(g(x)) \). This becomes \( g(x^2 - 3) = (x^2 - 3)^2 - 3 \). Expanding \((x^2 - 3)^2\) gives \( x^4 - 6x^2 + 9 \), so \( x^4 - 6x^2 + 9 - 3 = x^4 - 6x^2 + 6 \).

Key Concepts

Function Operations

Function operations involve combining functions in various ways. These operations are akin to arithmetic operations, where functions are added, subtracted, multiplied, or divided. By combining functions, you can create new functions with unique properties and applications.
Here are some key function operations:

• Addition: If you have two functions, say \( f(x) \) and \( g(x) \), then their sum, \( (f + g)(x) \), is defined as \( f(x) + g(x) \).
• Subtraction: The difference, \( (f - g)(x) \), is \( f(x) - g(x) \).
• Multiplication: The product, \( (f \cdot g)(x) \), is \( f(x) \cdot g(x) \).
• Division: The quotient, \( \left(\frac{f}{g}\right)(x) \), is \( \frac{f(x)}{g(x)} \); though, remember, \( g(x) eq 0 \).

These operations allow for the exploration of various functional expressions, providing insights into new functional relationships and their implications in different contexts.

Function Composition

Function composition involves combining two functions in such a way that the output of one function becomes the input of another. It is represented by the notation \( (f \circ g)(x) \), which reads as "f composed with g of x." To compute this, you substitute the entire function \( g(x) \) into the function \( f(x) \).
For example, if \( f(x) = x + 5 \) and \( g(x) = x^2 - 3 \), the composition \( f(g(x)) \) means you replace every \( x \) in \( f(x) \) with \( g(x) \). This gives:
\( f(g(x)) = f(x^2 - 3) = (x^2 - 3) + 5 = x^2 + 2 \).
Function composition allows you to build more complex functions from simpler ones, helping to model intricate scenarios in mathematical and real-world applications.

Domain and Range

The concepts of domain and range are foundational in understanding functions. They describe the valid inputs and outputs of a function, respectively.

• Domain: The domain of a function is the set of all possible input values (usually \( x \) values) that the function can accept without resulting in an undefined situation.


• Range: The range is the set of all possible output values (usually \( y \) values or \( f(x) \) results) that the function can produce.

When working with function compositions, understanding the domain becomes even more critical as the output range of the inner function must align with the domain of the outer function. This ensures that every composed expression leads to valid outcomes without errors or undefined values.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, typically expressed in the standard form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. These functions graph as parabolas on the coordinate plane.
Key characteristics of quadratic functions include:

• Vertex: The vertex is the highest or lowest point on the graph of a quadratic function.
• Axis of symmetry: This is a vertical line that divides the parabola into two mirror-image halves.
• Direction: The parabola opens upwards if \( a > 0 \) and downwards if \( a < 0 \).


• Intercepts: These are the points where the graph intersects the \( x \)-axis and \( y \)-axis.

Quadratic functions are widely used in various fields, from physics to economics, because they can represent scenarios involving area, projectile motion, and optimization problems.