Problem 16 - College Algebra and Calculus: An Applied Approach

Question

Evaluate the function for \(f(x)=2 x+1\) and \(g(x)=x^{2}-2\) \((f+g)(t-1)\)

Step-by-Step Solution

Answer

The evaluated function \( (f+g)(t-1) \) equals \( t^{2} \)

Step 1: Understand the function \( f(x) \)

The function \( f(x) \) is given as \( f(x) = 2x+1 \). This is a linear function with slope 2 and y-intercept 1.

Step 2: Understand the function \( g(x) \)

The function \( g(x) \) is given as \( g(x) = x^{2} - 2 \). This is a quadratic function with a negative constant term.

Step 3: Substitute values of \( t-1 \) into \( f(x) \) and \( g(x) \)

Substitute \( t-1 \) for \( x \) in both \( f(x) \) and \( g(x) \). For \( f(x) \), this gives \( f(t-1) = 2(t-1)+1 = 2t-2+1 = 2t-1 \). For \( g(x) \), this gives \( g(t-1) = (t-1)^{2} - 2 = t^{2} - 2t +1 - 2 = t^{2} - 2t - 1 \).

Step 4: Add the functions

Now, add \( f(t-1) \) and \( g(t-1) \) together to find \( (f+g)(t-1) \). This gives \( (f+g)(t-1) = (2t-1) + (t^{2} - 2t - 1) = t^{2} + 2t - 1 - 2t + 1 = t^{2} \).

Key Concepts

Linear Functions

Linear functions are the simplest type of functions, and they create straight lines when graphed on a coordinate plane. A general form for a linear function is \( f(x) = mx + b \), where \( m \) is the slope, and \( b \) is the y-intercept. The slope \( m \) represents how steep the line is, and the y-intercept \( b \) is the point where the line crosses the y-axis.

In the function \( f(x) = 2x + 1 \), the slope is 2, which means for every unit increase in \( x \), \( f(x) \) increases by 2 units. The y-intercept is 1, so the line crosses the y-axis at the point (0,1). In practical terms, linear functions describe relationships involving constant rates of change, such as speed or rate of pay.

When we substitute \( t-1 \) for \( x \) in this function, we perform algebraic operations to find \( f(t-1) \), illustrating how changes in inputs affect outputs in a straightforward manner.

Quadratic Functions

Quadratic functions are a step more complex than linear ones. They are shaped like a parabola on a graph, which can open upwards or downwards. The general form of a quadratic function is \( g(x) = ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \).

The quadratic function given, \( g(x) = x^2 - 2 \), lacks the linear \( bx \) term, simplifying our task. This parabola opens upward because the coefficient of \( x^2 \) is positive (1 in this case). The "-2" indicates a downward shift of the entire curve by 2 units.

Substituting \( t-1 \) for \( x \) helps us to evaluate the steeper curves that quadratic functions provide. Calculating \( g(t-1) \) requires expanding and simplifying, which showcases the quadratic relationship where the output isn't constant, but rather changes at an increasing rate as \( x \) changes.

Composite Functions

Composite functions involve combining two functions to form a new one. When adding functions, as in the problem, we're combining linear and quadratic functions to see how their interactions affect the result.

In our example, we evaluated and then added \( f(t-1) \) and \( g(t-1) \), which involved processing the outputs from each separate function and uniting them into a single expression. This process allows us to understand how different mathematical models (like linear and quadratic) can combine, resulting in richer, more complex behaviors.

Evaluating \( (f+g)(t-1) \) by substituting and simplifying gave us \( t^2 \). This result illustrates simplification through combining, where terms cancel out to leave a new function type springing from two different origins, specifically highlighting the ease with which mathematical operations on functions can reduce apparent complexity.