Chapter 5: Problem 12
Use the given pair of functions to find the following values if they exist. $$\bullet (g \circ f)(0)$$ $$\bullet (f \circ g)(-1)$$ $$\bullet (f \circ f)(2)$$ $$\bullet (g \circ f)(-3)$$ $$\bullet (f \circ g)\left(\frac{1}{2}\right)$$ $$\bullet (f \circ f)(-2)$$ $$f(x)=\sqrt{2 x+5}, g(x)=\frac{10 x}{x^{2}+1}$$
Short Answer
Step by step solution
Calculate \((g \circ f)(0)\)
Calculate \((f \circ g)(-1)\)
Calculate \((f \circ f)(2)\)
Calculate \((g \circ f)(-3)\)
Calculate \((f \circ g)\left(\frac{1}{2}\right)\)
Calculate \((f \circ f)(-2)\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Square Root Function
- 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.
Rational Function
- 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.
Domain of a Function
- 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.
Function Evaluation
- 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.