Chapter 5: Problem 47
\(f(x)=2 \sqrt{x}, g(x)=f(x+3)\)
Short Answer
Expert verified
The function \(g(x) = 2\sqrt{x+3}\).
Step by step solution
01
Identify the functions
The functions given in the exercise are \(f(x) = 2\sqrt{x}\) and \(g(x) = f(x+3)\).
02
Substitute \(x+3\) into \(f(x)\)
To find \(g(x)\), substitute \(x+3\) into \(f(x)\) replacing \(x\), which gives us \(f(x+3) = 2\sqrt{x+3}\).
03
Simplify \(g(x)\)
Therefore, \(g(x)\) can be simplified to \(g(x) = 2\sqrt{x+3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function Operations
Understanding function operations is essential for working with different types of functions in mathematics. Function operations include the addition, subtraction, multiplication, and division of functions, as well as function composition. In the case of the provided exercise, we are dealing with a special type of operation known as function composition.
Function composition involves creating a new function by applying one function to the results of another. This process is represented by \( g(x) = f(h(x)) \) where \( h(x) \) is another function applied to \( x \) before being passed on to the function \( f \) . The step by step solution shows the composite function with \( g(x) = f(x + 3) \) which indicates that the function \( f \) is applied to the expression \( x + 3 \) instead of just \( x \) .
To perform operations on functions accurately, it is important to first understand the individual functions involved and then correctly apply them according to the defined operation. With practice, manipulating these operations becomes more intuitive, allowing for a deeper exploration of function behavior.
Function composition involves creating a new function by applying one function to the results of another. This process is represented by \( g(x) = f(h(x)) \) where \( h(x) \) is another function applied to \( x \) before being passed on to the function \( f \) . The step by step solution shows the composite function with \( g(x) = f(x + 3) \) which indicates that the function \( f \) is applied to the expression \( x + 3 \) instead of just \( x \) .
To perform operations on functions accurately, it is important to first understand the individual functions involved and then correctly apply them according to the defined operation. With practice, manipulating these operations becomes more intuitive, allowing for a deeper exploration of function behavior.
Square Root Functions
Square root functions are a type of radical function that are represented in the form \( f(x) = \sqrt{x} \) or some variation thereof. They are distinctive due to their graph which forms a curve that starts at the origin (or a translated point) and extends infinitely to the right, graphically resembling half of a parabola turned sideways.
Handling a square root function requires understanding its domain and range. It is important to note that the square root of a number is only real-valued when the number under the square root is non-negative. Therefore, the domain of a basic square root function such as \( f(x) = \sqrt{x} \) is \( x \geq 0 \) .
The square root function is inherently related to the function \( f(x) = x^2 \) as the inverse operation; squaring is the process of multiplying a number by itself and the square root is finding the number that was squared to get the given number. This fundamental understanding helps when manipulating square root functions within more complex expressions and function operations.
Handling a square root function requires understanding its domain and range. It is important to note that the square root of a number is only real-valued when the number under the square root is non-negative. Therefore, the domain of a basic square root function such as \( f(x) = \sqrt{x} \) is \( x \geq 0 \) .
The square root function is inherently related to the function \( f(x) = x^2 \) as the inverse operation; squaring is the process of multiplying a number by itself and the square root is finding the number that was squared to get the given number. This fundamental understanding helps when manipulating square root functions within more complex expressions and function operations.
Function Substitution
Function substitution is a method used to evaluate composite functions, where one function is substituted into another. It is especially useful when dealing with exercises that involve finding the output of a function from a certain input value.
The exercise provided utilizes function substitution to find \( g(x) \) by substituting \( x+3 \) into the function \( f \) . It is a crucial step in evaluating composite functions and understanding how changes in the input affect the output.
To perform function substitution correctly, you must ensure that you've replaced every instance of the input variable with the expression you're substituting in. In the step-by-step solution, \( x \) in \( f(x) \) is completely replaced with \( x+3 \) , resulting in \( f(x+3) \) which simplifies to \( 2\sqrt{x+3} \) .
This concept becomes more intuitive over time with practice and can be applied to various types of functions, enabling the solution of a wide range of problems in algebra and calculus. Substitution not only aids in evaluating composite functions but also in understanding the intricate relationship between multiple functions when one is nested within the other.
The exercise provided utilizes function substitution to find \( g(x) \) by substituting \( x+3 \) into the function \( f \) . It is a crucial step in evaluating composite functions and understanding how changes in the input affect the output.
To perform function substitution correctly, you must ensure that you've replaced every instance of the input variable with the expression you're substituting in. In the step-by-step solution, \( x \) in \( f(x) \) is completely replaced with \( x+3 \) , resulting in \( f(x+3) \) which simplifies to \( 2\sqrt{x+3} \) .
This concept becomes more intuitive over time with practice and can be applied to various types of functions, enabling the solution of a wide range of problems in algebra and calculus. Substitution not only aids in evaluating composite functions but also in understanding the intricate relationship between multiple functions when one is nested within the other.