Problem 73 - Algebra 2

Question

Let \(f(x)=x^{2}+1\) and \(g(x)=3 x .\) Find each value. \((g \circ f)\left(\frac{1}{2}\right)\)

Step-by-Step Solution

Answer

The final result is \(15/4\).

Step 1: Calculate \(f(x)\) for \(x = 1/2\)

We substitute \(x = 1/2\) into \(f(x)\) to get \(f(1/2)=(1/2)^{2}+1=1/4+1=5/4.\)

Step 2: Substitute \(f(1/2)\) into \(g(x)\)

Now we take the result from the first step (5/4) and we substitute it into \(g(x)\) to get \(g(f(1/2)) = g(5/4) = 3*(5/4) = 15/4.\)

Key Concepts

Quadratic Function

A quadratic function is a polynomial function that can be expressed in the form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Typically, quadratic functions are represented by a parabola when graphed on a coordinate plane. The standard form makes it easy to identify the key components of the quadratic:

• Vertex: The highest or lowest point of the parabola, depending on its orientation.
• Axis of Symmetry: The vertical line that passes through the vertex and divides the parabola into two mirror images.
• Roots/Zeros: The points where the parabola crosses the x-axis, if they exist.
• Direction: Determined by the sign of \( a \); if \( a > 0 \), the parabola opens upwards, and if \( a < 0 \), it opens downwards.

Given the exercise, the quadratic function \( f(x) = x^2 + 1 \) is in its simplest form without linear or constant adjustments apart from \( c = 1 \). This means it is an upright parabola with its vertex at (0, 1), as there is no \( x \) and \( ax^2 \) leads. Evaluating quadratic functions is fundamental when tackling problems involving function compositions like in our given task.

Linear Function

A linear function is defined by the equation \( g(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. The slope \( m \) indicates how steep the line is, and the y-intercept \( b \) is the point where the line crosses the y-axis. Linear functions are characterized by their straight-line graphs.

• These functions have a constant rate of change, making them predictable and simple compared to nonlinear functions like quadratics.
• The slope \( m \) can be calculated if two points on the line are known, using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \).

In our context, the linear function \( g(x) = 3x \) has a slope of 3 and no constant \( b \), which means it passes through the origin (0, 0). Linear functions often offer simplicity when used in compositions alongside quadratic functions, as seen in evaluating \( g(f(x)) \). This is crucial to understanding how the constant multiplier affects our solution when applying the result of \( f(x) \) to \( g(x) \).

Function Evaluation

Function evaluation involves calculating the output of a function given a specific input. It means substituting the given input value into the function expression and performing any necessary algebraic operations.

• In the process of function evaluation, consider whether the function is linear, quadratic, or of another type. This can affect the steps required for computation.
• Function compositions, like \((g \circ f)(x)\), involve evaluating one function and then using its result as the input for another function. This can be broken down into distinct steps to simplify the process.

In the exercise, we first evaluate the function \( f(x) \) at \( x = \frac{1}{2} \) to get \( f(\frac{1}{2}) = \frac{5}{4} \). Afterward, this result becomes the input for \( g(x) \) in the composition. This systematic approach showcases how nested functions interact through evaluation. Mastery of this concept provides a strong foundation for solving complex function-based problems.