Problem 20 - Contemporary Precalculus

Question

Find \((g \circ f)(3),(f \circ g)(1),\) and \((f \circ f)(0)\) $$f(x)=x^{2}-1, \quad g(x)=\sqrt{x}$$

Step-by-Step Solution

Answer

Answer: The compositions of the given functions are: - \((g \circ f)(3) = \sqrt{8}\) - \((f \circ g)(1) = 0\) - \((f \circ f)(0) = 0\)

Step 1: Find \((g \circ f)(3)\)

To find \((g \circ f)(3)\), we need to substitute \(f(3)\) into the function \(g(x)\): $$g(f(3)) = g(3^{2} - 1)$$ Now, calculate \(3^2 - 1 = 8\), and substitute into the function \(g\): $$g(8) = \sqrt{8}$$ So, \((g \circ f)(3) = \sqrt{8}\).

Step 2: Find \((f \circ g)(1)\)

To find \((f \circ g)(1)\), we need to substitute \(g(1)\) into the function \(f(x)\): $$f(g(1)) = f(\sqrt{1})$$ Now, calculate \(\sqrt{1} = 1\), and substitute into the function \(f\): $$f(1) = 1^{2} -1$$ So, \((f \circ g)(1) = 1^2 - 1 = 0\).

Step 3: Find \((f \circ f)(0)\)

To find \((f \circ f)(0)\), we need to substitute \(f(0)\) into the function \(f(x)\) again: $$f(f(0)) = f(0^{2} - 1)$$ Now, calculate \(0^2 - 1 = -1\), and substitute into the function \(f\): $$f(-1) = (-1)^{2} - 1$$ So, \((f \circ f)(0) = (-1)^2 - 1 = 0\). In summary, the compositions are: - \((g \circ f)(3) = \sqrt{8}\) - \((f \circ g)(1) = 0\) - \((f \circ f)(0) = 0\)

Key Concepts

Composite Functions

Composite functions combine two or more functions to create a new function. They are denoted by the symbol \( \circ \), which means "of." In the expression \((g \circ f)(x)\), we apply \(f\) first and then \(g\) second. This might seem tricky at first, but with practice, it becomes a straightforward process.

To solve composite functions, follow these steps:

• Identify the inner and outer functions. In \((g \circ f)(x)\), \(f\) is the inner function and \(g\) is the outer function.
• Evaluate the inner function at the given value \(x\).
• Substitute the result into the outer function.

For example, to find \((g \circ f)(3)\), first evaluate \(f(3)\), which is \(3^2 - 1 = 8\). Next, substitute into \(g\), so \(g(8) = \sqrt{8}\). The result is \(\sqrt{8}\).

By splitting the problem into these steps, complex function composition becomes more manageable. Remember, practice makes perfect!

Square Roots

Square roots are a type of radical that helps us find a number which, when multiplied by itself, gives the original number. The square root symbol, \(\sqrt{}\), denotes this operation, and understanding square roots is crucial in many areas of mathematics.

Key things to remember about square roots include:

• They only apply to non-negative numbers in the real number system since the square root of a negative number isn't a real number.
• The square root of a perfect square results in an integer, like \(\sqrt{4} = 2\) or \(\sqrt{9} = 3\).
• For non-perfect squares, the result is often an irrational number, like \(\sqrt{8}\).

Square roots frequently appear in problems involving functions, such as \(g(x) = \sqrt{x}\). Evaluating \(g(x)\) requires computing the square root of the input value. An understanding of how to handle both perfect and non-perfect squares will make these tasks easier.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, and they usually have the general form \(f(x) = ax^2 + bx + c\). In our exercise, the quadratic function is \(f(x) = x^2 - 1\).

Here are some important features of quadratic functions:

• The graph of a quadratic function is a parabola, which can open upwards or downwards depending on the sign of \(a\).
• They have a vertex that represents the highest or lowest point of the parabola.
• The solutions or "roots" of the quadratic are values of \(x\) where the function equals zero, found by setting \(f(x) = 0\) and solving for \(x\).

In our problem, substituting values into the quadratic function \(f(x) = x^2 - 1\) helps us find how it transforms other function outputs. For instance, substituting \(x = 3\) gives \(f(3) = 3^2 - 1 = 8\). When combined with another function, such as a square root, this can solve composite function problems like \((g \circ f)(x)\).

Understanding quadratic functions and their properties is essential for solving a wide array of mathematical problems.