Problem 20
Question
(a) find the value of the function at \(x=0, x=1\), and \(x=-1\). (b) find all \(x\) such that the value of the function is \((i) 0,(i i) 1\), and \((i i i)-1 .\) $$ g(x)=x^{2}-1 $$
Step-by-Step Solution
Verified Answer
For part a, the function values are: \(g(0) = -1\), \(g(1) = 0\) and \(g(-1) = 0\). For part b, the values for \(x\) when the function is 0 are \(-1\) and \(1\), when the function is 1 they are \(-\sqrt{2}\) and \(\sqrt{2}\), and when the function is -1 it is \(0\).
1Step 1: Find the value of function at given points
Substitute \(x=0, x=1, x=-1\) in the function \(g(x)=x^{2}-1\) to find the value of the function at these points.-> \(g(0)=(0)^{2}-1=-1\)-> \(g(1)=(1)^{2}-1=0\)-> \(g(-1)=(-1)^{2}-1=0\)
2Step 2: Find all x such that the value of the function is 0
To find the \(x\) value when the function is zero, set \(g(x)=0\). This yields the equation \(x^{2}-1=0\).Solving for \(x\) we get, \(x^{2}=1\), which gives \(x=1\) and \(x=-1\)
3Step 3: Find all x such that the value of the function is 1
To find the \(x\) value when the function is one, set \(g(x)=1\). This yields the equation \(x^{2}-1=1\).Solving for \(x\) we get, \(x^{2}=2\), which gives \(x= \sqrt{2}\) and \(x= -\sqrt{2}\)
4Step 4: Find all x such that the value of the function is -1
To find the \(x\) value when the function is negative one, set \(g(x)=-1\). This yields the equation \(x^{2}-1=-1\).After solving for \(x\) we get, \(x^{2}=0\), which gives \(x=0\)
Key Concepts
Function EvaluationFind x Given Function ValueSolving for xQuadratic Function Properties
Function Evaluation
Function evaluation is a fundamental concept in algebra which involves finding the output of a function given an input value. In our exercise, the function given is g(x) = x^2 - 1, and we're tasked with evaluating this function at different values of x—specifically, x=0, x=1, and x=-1.
To evaluate the function at these points, we simply substitute the value of x into the function. For example, when x=0, we compute g(0) = (0)^2 - 1 = -1. This process is repeated for the other values of x. Being able to accurately evaluate a function at specific points helps us to understand the function’s behavior and is a crucial step in graphing or analyzing a function's properties.
To evaluate the function at these points, we simply substitute the value of x into the function. For example, when x=0, we compute g(0) = (0)^2 - 1 = -1. This process is repeated for the other values of x. Being able to accurately evaluate a function at specific points helps us to understand the function’s behavior and is a crucial step in graphing or analyzing a function's properties.
Find x Given Function Value
Determining the x-values that correspond to a specific function value is a key skill in algebra. It involves working backwards: we start with the function's output and seek the input(s) that would produce it. In our case, g(x) = x^2 - 1, and we’re looking for all x such that g(x) equals 0, 1, and -1.
Setting g(x) equal to these values leads to solving quadratic equations. For example, when finding x such that g(x) = 0, we get the equation x^2 - 1 = 0. By factoring or using the square root principle, we can find the values of x that satisfy this equation (x = 1 and x = -1). This method applies to other given function values as well, such as g(x) = 1 or g(x) = -1, leading us to the respective x-values by solving the resulting equations.
Setting g(x) equal to these values leads to solving quadratic equations. For example, when finding x such that g(x) = 0, we get the equation x^2 - 1 = 0. By factoring or using the square root principle, we can find the values of x that satisfy this equation (x = 1 and x = -1). This method applies to other given function values as well, such as g(x) = 1 or g(x) = -1, leading us to the respective x-values by solving the resulting equations.
Solving for x
The process of solving for x requires manipulating an equation to isolate the variable x on one side. In quadratic equations, like the ones derived from our function g(x) = x² - 1, we often find two solutions for x because a quadratic equation can intersect the x-axis at two points, or in the case of a perfect square, touch it at just one point.
The solutions to a quadratic equation can be found using factoring, completing the square, or applying the quadratic formula. In our exercise, we used factoring and the square root method to solve the equations obtained by setting the function equal to different values. For instance, to solve for x when g(x) = 0, we factored the equation to (x + 1)(x - 1) = 0, leading to the solutions x = 1 and x = -1. Each method for solving x is like a tool in a toolbox; you choose the best one depending on the equation you're working with.
The solutions to a quadratic equation can be found using factoring, completing the square, or applying the quadratic formula. In our exercise, we used factoring and the square root method to solve the equations obtained by setting the function equal to different values. For instance, to solve for x when g(x) = 0, we factored the equation to (x + 1)(x - 1) = 0, leading to the solutions x = 1 and x = -1. Each method for solving x is like a tool in a toolbox; you choose the best one depending on the equation you're working with.
Quadratic Function Properties
Quadratic functions, such as g(x) = x² - 1 from our exercise, have several distinctive properties. One of them is symmetry; the graph of a quadratic function is a parabola that opens upward or downward and is symmetrical about a vertical line called the axis of symmetry.
Another property is the vertex, which is the highest or lowest point on the graph, depending on whether the parabola opens downwards or upwards, respectively. In the case of g(x), the vertex is (0, -1), and the parabola opens upwards since the coefficient of x² is positive. Quadratic functions also have up to two real roots — the values of x where g(x) = 0. These are the points where the graph intersects the x-axis, and in this exercise, they are x = 1 and x = -1. Understanding these properties helps us to visualize functions and find solutions to equations more intuitively.
Another property is the vertex, which is the highest or lowest point on the graph, depending on whether the parabola opens downwards or upwards, respectively. In the case of g(x), the vertex is (0, -1), and the parabola opens upwards since the coefficient of x² is positive. Quadratic functions also have up to two real roots — the values of x where g(x) = 0. These are the points where the graph intersects the x-axis, and in this exercise, they are x = 1 and x = -1. Understanding these properties helps us to visualize functions and find solutions to equations more intuitively.
Other exercises in this chapter
Problem 17
If \(j(x)=3 x^{2}-2 x+1\), find the following. Simplify your answer where possible. (a) \(j(0)\) (b) \(j(1)\) (c) \(j(-1)\) (d) \(j(-x)\) (e) \(j(x+2)\) (f) \(3
View solution Problem 19
In Problems 19 through 21 (a) find the value of the function at \(x=0, x=1\), and \(x=-1\). (b) find all \(x\) such that the value of the function is \((i) 0,(i
View solution Problem 21
(a) find the value of the function at \(x=0, x=1\), and \(x=-1\). (b) find all \(x\) such that the value of the function is \((i) 0,(i i) 1\), and \((i i i)-1 .
View solution Problem 22
For each function in Problems 22 through 27, determine the largest possible domain. (a) \(f(x)=\frac{1}{5 x+10}\) (b) \(g(x)=\sqrt{5 x+10}\)
View solution