Problem 12 - Precalculus

Question

Use the given pair of functions to find the following values if they exist. $$\bullet (g \circ f)(0)$$ $$\bullet (f \circ g)(-1)$$ $$\bullet (f \circ f)(2)$$ $$\bullet (g \circ f)(-3)$$ $$\bullet (f \circ g)\left(\frac{1}{2}\right)$$ $$\bullet (f \circ f)(-2)$$ $$f(x)=\sqrt{2 x+5}, g(x)=\frac{10 x}{x^{2}+1}$$

Step-by-Step Solution

Answer

\((g \circ f)(0) = \frac{5\sqrt{5}}{3}\), \((f \circ g)(-1)\) does not exist, \((f \circ f)(2) = \sqrt{11}\), \((g \circ f)(-3)\) does not exist, \((f \circ g)\left(\frac{1}{2}\right) = \sqrt{13}\), \((f \circ f)(-2) = \sqrt{7}\).

Step 1: Calculate \((g \circ f)(0)\)

To find \((g \circ f)(0)\), first calculate \(f(0)\).Given \(f(x) = \sqrt{2x + 5}\), substitute \(x = 0\): \[ f(0) = \sqrt{2(0) + 5} = \sqrt{5} \]Then evaluate \(g(\sqrt{5})\) using \(g(x) = \frac{10x}{x^2 + 1}\).Substitute \(x = \sqrt{5}\):\[ g(\sqrt{5}) = \frac{10(\sqrt{5})}{(\sqrt{5})^2 + 1} = \frac{10\sqrt{5}}{5 + 1} = \frac{10\sqrt{5}}{6} = \frac{5\sqrt{5}}{3} \]

Step 2: Calculate \((f \circ g)(-1)\)

First, evaluate \(g(-1)\).Given \(g(x) = \frac{10x}{x^2 + 1}\), substitute \(x = -1\):\[ g(-1) = \frac{10(-1)}{(-1)^2 + 1} = \frac{-10}{1 + 1} = \frac{-10}{2} = -5 \]Next, evaluate \(f(-5)\) using \(f(x) = \sqrt{2x + 5}\).Substitute \(x = -5\):\[ f(-5) = \sqrt{2(-5) + 5} = \sqrt{-10 + 5} = \sqrt{-5} \]Since the square root of a negative number is not a real number, \(f(-5)\) does not exist, and hence \((f \circ g)(-1)\) does not exist.

Step 3: Calculate \((f \circ f)(2)\)

First, evaluate \(f(2)\).Given \(f(x) = \sqrt{2x + 5}\), substitute \(x = 2\):\[ f(2) = \sqrt{2(2) + 5} = \sqrt{4 + 5} = \sqrt{9} = 3 \]Then find \(f(3)\).Substitute \(x = 3\):\[ f(3) = \sqrt{2(3) + 5} = \sqrt{6 + 5} = \sqrt{11} \]

Step 4: Calculate \((g \circ f)(-3)\)

First, calculate \(f(-3)\).Given \(f(x) = \sqrt{2x + 5}\), substitute \(x = -3\):\[ f(-3) = \sqrt{2(-3) + 5} = \sqrt{-6 + 5} = \sqrt{-1} \]Since the square root of a negative number is not a real number, \(f(-3)\) does not exist, and hence \((g \circ f)(-3)\) does not exist.

Step 5: Calculate \((f \circ g)\left(\frac{1}{2}\right)\)

Start by finding \(g\left(\frac{1}{2}\right)\).Given \(g(x) = \frac{10x}{x^2 + 1}\), substitute \(x = \frac{1}{2}\):\[ g\left(\frac{1}{2}\right) = \frac{10 \left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)^2 + 1} = \frac{5}{\frac{1}{4} + 1} = \frac{5}{\frac{5}{4}} = 4 \]Then compute \(f(4)\).Given \(f(x) = \sqrt{2x + 5}\),\[ f(4) = \sqrt{2(4) + 5} = \sqrt{8 + 5} = \sqrt{13} \]

Step 6: Calculate \((f \circ f)(-2)\)

Find \(f(-2)\) first.Given \(f(x) = \sqrt{2x + 5}\), substitute \(x = -2\):\[ f(-2) = \sqrt{2(-2) + 5} = \sqrt{-4 + 5} = \sqrt{1} = 1 \]Then compute \(f(1)\).Given \(f(x) = \sqrt{2x + 5}\), substitute \(x = 1\):\[ f(1) = \sqrt{2(1) + 5} = \sqrt{2 + 5} = \sqrt{7} \]

Key Concepts

Square Root Function

The square root function is a mathematical operation that finds the number that, when multiplied by itself, equals the given number. It is denoted by the symbol \( \sqrt{} \). The function typically looks like \( f(x) = \sqrt{x} \), and takes only non-negative values under the square root. This is crucial because the square root of a negative number is not part of the set of real numbers, but rather imaginary.

• Understanding: Ensure that the expression inside the square root (called the radicand) is non-negative if we want a real output. For example, in the function \( f(x) = \sqrt{2x + 5} \), the condition for the function to have real outputs is \( 2x + 5 \geq 0 \).
• Applications: Square root functions are used widely in various fields of science and mathematics to describe relationships involving area, speed, and statistics.

An important aspect in our context is remembering that if you have to deal with negative radicands, like \( \sqrt{-1} \), then the solution does not exist within the set of real numbers. Therefore, for \( f(x) = \sqrt{2x + 5} \), make sure \( x \) satisfies the condition \( 2x + 5 \geq 0 \).

Rational Function

A rational function is a type of function represented by the ratio of two polynomials. It has the form \( g(x) = \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials, and \( q(x) eq 0 \). This is because division by zero is undefined within real numbers.

• Key Aspects: The denominator of a rational function cannot be zero. Hence, always identify values of \( x \) which might make the denominator zero, as these are not included in the function’s domain.
• Example: With \( g(x) = \frac{10x}{x^2 + 1} \), this is a rational function. No real number will make the denominator \( x^2 + 1 = 0 \), hence the domain is all real numbers.

Rational functions often describe real-world situations where one quantity depends on another in terms of rate, and they are extensively used in fields like economics, biology, and engineering.

Domain of a Function

The domain of a function is the complete set of possible values of the independent variable, usually represented as \( x \). It determines what values can be input into the function without causing undefined behavior (such as division by zero or taking the square root of a negative number).

• Finding the Domain:
  • For square root functions like \( f(x) = \sqrt{2x + 5} \), the expression inside the square root must be non-negative, giving \( 2x + 5 \geq 0 \). Simplifying, \( x \geq -\frac{5}{2} \).
  • For rational functions, solve for where the denominator equals zero to find exclusion points.
• Validation: Always check computation via substitution to ensure exclusion point correctness.

Understanding the domain helps avoid errors and ensures that the function outputs realistic and expected results. This is fundamental when dealing with function composition.

Function Evaluation

Function evaluation involves substituting a specific value for \( x \) in the function and determining the output. This process helps in understanding how the function behaves and determining specific outcomes based on particular inputs.

• Calculation: To evaluate a function, follow mathematical operations as dictated by the function’s definition. For instance, if \( f(x) = \sqrt{2x + 5} \), to find \( f(0) \), substitute and solve: \( \sqrt{2(0) + 5} = \sqrt{5} \).
• Complex Evaluations: Involves multi-step evaluations like evaluating \( f(g(x)) \) or \( g(f(x)) \). Start from the inside function, then use the result as input for the outer function.

Proper function evaluation is key in function composition. Missteps in single function evaluation could cascade into errors when dealing with compositions. Always approach step-by-step, checking each outcome before proceeding to ensure accuracy.