Problem 57 - Precalculus : Building Concepts and Connections

Question

In Exercises \(49-66,\) let \(f(x)=x^{2}+x, g(x)=\sqrt{x},\) and \(h(x)=-3 x\) Evaluate each of the following. $$(f \circ g)(4)$$

Step-by-Step Solution

Answer

The value of \( (f \circ g)(4) \) is 6.

Step 1: Find \(g(4)\)

Since \(g(x)=\sqrt{x}\), substitute x with 4 to find \(g(4)\). So, \(g(4)=\sqrt{4}\) which equals 2.

Step 2: Find \(f(g(4))\)

Now that you have found \(g(4)\) to be 2, substitute this value into the function \(f(x)\) to find \(f(g(4))\). So \(f(g(4)) = f(2) = 2^2 + 2 = 6\).

Step 3: Conclude

From the previous steps, you can see that \( (f \circ g)(4) = f(g(4)) = f(2) = 6\).

Key Concepts

Composite Functions

A composite function is created when two functions are combined in such a way that the output of one function becomes the input for another. In mathematical notation, this is represented as \((f \circ g)(x) = f(g(x))\). This tells us to first apply the function \(g\) to \(x\), and then use the result as the input for the function \(f\).

Composite functions are useful because they allow complex operations to be broken down into simpler parts. They are widely used in various fields, such as calculus and real-world applications, to create more sophisticated models that can solve complicated problems.

• Break down complex computations into manageable steps.
• Combine simple functions to create complex transformations.

Function Evaluation

Function evaluation involves finding the output value of a function for a particular input. Let's look at how we evaluate composite functions through an example.

If given \(f(x) = x^2 + x\), \(g(x) = \sqrt{x}\), and you need to find \((f \circ g)(4)\):

• Evaluate \(g(4)\) first: Since \(g(x) = \sqrt{x}\), replace \(x\) with 4. Thus, \(g(4) = \sqrt{4} = 2\).
• Next, use this result in function \(f\): Find \(f(2)\), which is \(2^2 + 2 = 4 + 2 = 6\).

Function evaluation, therefore, provides the computational pathway from input to output, allowing students to see how functions interact within composite operations.

Square Root Function

A square root function is one of the principal types of functions explored in mathematics. It is usually in the form \(g(x) = \sqrt{x}\). The domain of \(g(x)\) includes all non-negative numbers since a square root is not defined for negative numbers within the set of real numbers.

Some key points about the square root function:

• The square root of a number \(x\) is a value that, when multiplied by itself, gives \(x\).
• The output of a square root function is always non-negative.
• Simplifies calculations by reducing the index power.

In the composite function exercise, the square root function was used as \(g(x)\), where it takes a value and reduces it to its root, effectively transforming the input to be used in another function.

Quadratic Function

The quadratic function is represented by \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants with \(a eq 0\). It is one of the fundamental types of polynomial functions.

In our example, \(f(x) = x^2 + x\) served as the quadratic function. Here's what makes quadratic functions important:

• Their graph is a parabola, which can open upwards or downwards depending on the sign of \(a\).
• Quadratic functions can be used to model situations involving acceleration, area, and various natural phenomena.
• They involve operations like squaring and addition, making them excellent candidates for demonstrating the effects of different function inputs.

For this exercise, the quadratic function was applied to the result from \(g(x)\) to find \(f(2)\), showcasing their role in determining outputs based on composite functions.