/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 In Exercises \(49-66,\) let \(f(... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercises \(49-66,\) let \(f(x)=x^{2}+x, g(x)=\sqrt{x},\) and \(h(x)=-3 x\) Evaluate each of the following. $$(g \circ h)(-12)$$

Short Answer

Expert verified
The value of \(g \circ h(-12)\) is \(6\).

Step by step solution

01

Substituting \(h(-12)\) in function \(h\)

Find the value of function \(h(x)\) when \(x = -12\), this is denoted as \(h(-12)\):\nGiven that \(h(x)=-3x\), when \(x=-12\), the function becomes \(h(-12)= -3(-12)\).
02

Calculating the Value of \(h(-12)\)

Compute the multiplicative expression of the function to find the value of \(h(-12)\):\n\(h(-12) = 36\).
03

Substituting \(h(-12)\) in function \(g\)

Next, replace the value of \(h(-12)\) into function \(g(x)\) to get \(g(h(-12))\), which can also be denoted as \(g \circ h(-12)\):\nGiven that \(g(x) = \sqrt{x}\), our expression becomes \(g(36)\).
04

Computing the Value of \(g \circ h(-12)\)

Finally, compute the square root of 36 to find the result of \(g \circ h(-12)\):\n\(g \circ h(-12) = \sqrt{36} = 6\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
In precalculus, function evaluation is a fundamental concept, which involves finding the output of a function for a particular input value. To evaluate a function, you replace the variable in the function's rule (usually represented by 'x') with the given input value.

For example, if you have a function defined as f(x) = x2 + x, and you want to evaluate it at x = 2, you would substitute 2 for x in the expression, which yields f(2) = 22 + 2 = 4 + 2 = 6. This process gives you the output of the function for that specific input.

Function evaluation is not only crucial for simple functions but also forms the backbone of more complex operations involving functions, such as in the construction of composite functions.
Composite Functions
Composite functions involve one function taking another function as its input. The notation (g \(circ\) h)(x) symbolizes a composite function where function g is applied to the result of function h for an input x. The operation \(circ\) is known as function composition.

To calculate a composite function such as (g \(circ\) h)(-12), you start by evaluating the inner function h at x = -12, which gives you a certain value. Then you take that resulting value and use it as the input for the outer function g. This multi-step process is critical in complex problem-solving and creates new functions with their unique set of characteristics.
Square Root Function
The square root function is a type of radical function, specifically defined as g(x) = \(sqrt{x\) for x greater than or equal to zero. This function takes a non-negative input and gives its principal square root as the output.

In the context of the exercise, when you encounter g(36), you are essentially looking for the square root of 36. By applying the square root function, you get \(g(36) = \(sqrt{36}\) = 6\). Understanding how square root functions work is fundamental when dealing with quadratic equations and geometric applications involving areas or volumes.
Inverse Operations
Inverse operations in mathematics are pairs of operations that undo each other. For instance, addition and subtraction are inverses, as are multiplication and division. This concept is significantly extended in functions where an inverse function essentially reverses the effect of the original function.

If a function f maps an input x to an output y, then, if f is invertible, its inverse f-1 will map y back to x. This is notably true for the square function and the square root function. The square root is the inverse operation of squaring, as (\(sqrt{x}\))2 = x, for all non-negative x. Grasping inverse operations is beneficial when solving equations and in understanding how different functions relate to one another.

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Most popular questions from this chapter

Attendance at Broadway shows in New York can be modeled by the quadratic function \(p(t)=0.0489 t^{2}-0.7815 t+10.31,\) where \(t\) is the number of years since 1981 and \(p(t)\) is the attendance in millions of dollars. The model is based on data for the years \(1981-2000 .\) When did the attendance reach \(\$ 12\) million? (Source: The League of American Theaters and Producers, Inc.)

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In Exercises \(101-104,\) let \(f(t)=3 t+1\) and \(g(x)=x^{2}+4\). Evaluate \((g \circ g)\left(\frac{1}{2}\right)\)

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