Problem 98
Question
In Exercises \(97-100,\) let \(f(t)=-t^{2}\) and \(g(x)=x^{2}-1\). Evaluate \((g \circ g)\left(\frac{2}{3}\right)\)
Step-by-Step Solution
Verified Answer
\((g \circ g)\left(\frac{2}{3}\right) = -\frac{56}{81}\
1Step 1: Inserting the Value into the Function
We first insert \(\frac{2}{3}\) into function \(g(x)\). Therefore, \(g\left(\frac{2}{3}\right)=\left(\frac{2}{3}\right)^{2} - 1 = \frac{4}{9} - 1 = -\frac{5}{9}\)
2Step 2: Applying the Composition of Functions
Next, we insert the result \(-\frac{5}{9}\) back into function \(g\). This gives us \(g(g\left(\frac{2}{3}\right))=g\left(-\frac{5}{9}\right)=\left(-\frac{5}{9}\right)^{2}-1 = \frac{25}{81} - 1 = -\frac{56}{81}\)
3Step 3: Simplifying the Result
Finally, we simplify our result to the simplest form, and that gives us: \(-\frac{56}{81}\)
Key Concepts
Quadratic FunctionsFunction EvaluationAlgebraic Simplification
Quadratic Functions
Quadratic functions are a type of polynomial function that can be written in the form \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(aeq 0\). These functions create a parabolic shape when graphed, opening upwards if \(a > 0\) and downwards if \(a < 0\).
For this exercise, we worked with the function \(g(x) = x^2 - 1\). Here, the function simplifies to just the quadratic term minus one, which keeps the standard parabolic shape but shifts the graph down by one unit.
Quadratic functions are essential in many areas of mathematics and often appear in physics and engineering problems where projectile motion is considered. Understanding how to manipulate these functions helps in analyzing scenarios where acceleration due to gravity or other quadratic relationships occur.
For this exercise, we worked with the function \(g(x) = x^2 - 1\). Here, the function simplifies to just the quadratic term minus one, which keeps the standard parabolic shape but shifts the graph down by one unit.
Quadratic functions are essential in many areas of mathematics and often appear in physics and engineering problems where projectile motion is considered. Understanding how to manipulate these functions helps in analyzing scenarios where acceleration due to gravity or other quadratic relationships occur.
Function Evaluation
Function evaluation is the process of finding the output of a function for a specific input. For any given input value, you substitute it into the function equation to determine the corresponding output.
In our problem, we have the function \(g(x) = x^2 - 1\). To evaluate it for a specific input, such as \(x = \frac{2}{3}\), you substitute \(\frac{2}{3}\) into the function in place of \(x\). This results in computing \(\left(\frac{2}{3}\right)^2 - 1 = -\frac{5}{9}\).
Function evaluation is a fundamental skill in mathematics, as it allows you to understand what the function does to particular inputs. Whether you are dealing with real-world data or theoretical problems, evaluating functions accurately is crucial in generating reliable outcomes.
In our problem, we have the function \(g(x) = x^2 - 1\). To evaluate it for a specific input, such as \(x = \frac{2}{3}\), you substitute \(\frac{2}{3}\) into the function in place of \(x\). This results in computing \(\left(\frac{2}{3}\right)^2 - 1 = -\frac{5}{9}\).
Function evaluation is a fundamental skill in mathematics, as it allows you to understand what the function does to particular inputs. Whether you are dealing with real-world data or theoretical problems, evaluating functions accurately is crucial in generating reliable outcomes.
Algebraic Simplification
Algebraic simplification involves reducing expressions to their simplest and most concise form. It is a crucial step in mathematics, enabling easier interpretation and calculation of expressions.
In the exercise, after evaluating the composition of functions, we arrived at an expression that needed simplification: \(-\frac{56}{81}\). Here, algebraic simplification ensures the expression is as straightforward as possible, often by combining like terms or reducing fractions.
For fractions, simplification involves finding the greatest common factor of the numerator and the denominator to ensure the fraction is in its simplest form. Simplifying expressions is not only about making math easier to read but also about ensuring the solutions are in a form that is easily comparable to other results.
In the exercise, after evaluating the composition of functions, we arrived at an expression that needed simplification: \(-\frac{56}{81}\). Here, algebraic simplification ensures the expression is as straightforward as possible, often by combining like terms or reducing fractions.
For fractions, simplification involves finding the greatest common factor of the numerator and the denominator to ensure the fraction is in its simplest form. Simplifying expressions is not only about making math easier to read but also about ensuring the solutions are in a form that is easily comparable to other results.
Other exercises in this chapter
Problem 96
In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=(3 x-7)^{10}+5(3 x-7)^{2}$$
View solution Problem 97
In Exercises \(97-100,\) let \(f(t)=-t^{2}\) and \(g(x)=x^{2}-1\). Evaluate \((f \circ f)(-1)\)
View solution Problem 99
Can you write down an expression for a quadratic function whose \(x\) -intercepts are given by (2,0) and (3,0)\(?\) Is there more than one possible answer? Expl
View solution Problem 99
In Exercises \(97-100,\) let \(f(t)=-t^{2}\) and \(g(x)=x^{2}-1\). Find an expression for \((f \circ f)(t),\) and give the domain of \(f \circ f\).
View solution