Problem 50
Question
For the following exercises, given each function \(f\) , evaluate \(f(-1), f(0), f(2),\) and \(f(4)\) $$ f(x)=\left\\{\begin{array}{ll}{x^{2}-2} & {\text { if } x<2} \\ {4+|x-5|} & {\text { if } x \geq 2}\end{array}\right. $$
Step-by-Step Solution
Verified Answer
\( f(-1) = -1, f(0) = -2, f(2) = 7, f(4) = 5 \).
1Step 1: Understanding the Function
The function \( f(x) \) is defined piecewise. It has two parts: one for \( x < 2 \) where \( f(x) = x^2 - 2 \) and another for \( x \geq 2 \) where \( f(x) = 4 + |x - 5| \). We will use these definitions to evaluate the function at the given points.
2Step 2: Evaluate \( f(-1) \)
Since \( -1 < 2 \), we use the first part of the piecewise function: \( f(x) = x^2 - 2 \). Thus, \( f(-1) = (-1)^2 - 2 = 1 - 2 = -1 \).
3Step 3: Evaluate \( f(0) \)
Since \( 0 < 2 \), we use the first part of the piecewise function: \( f(x) = x^2 - 2 \). Thus, \( f(0) = 0^2 - 2 = 0 - 2 = -2 \).
4Step 4: Evaluate \( f(2) \)
Since \( 2 \geq 2 \), we use the second part of the piecewise function: \( f(x) = 4 + |x - 5| \). Thus, \( f(2) = 4 + |2 - 5| = 4 + 3 = 7 \).
5Step 5: Evaluate \( f(4) \)
Since \( 4 \geq 2 \), we use the second part of the piecewise function: \( f(x) = 4 + |x - 5| \). Thus, \( f(4) = 4 + |4 - 5| = 4 + 1 = 5 \).
Key Concepts
Function EvaluationAbsolute ValueQuadratic Functions
Function Evaluation
When dealing with functions, one of the fundamental tasks is to evaluate the function at given points. Function evaluation involves substituting a specific value of the variable into the function and calculating the result. For piecewise functions like the one given, this process can work in two steps: recognizing which part of the function applies to the given point and calculating the output accordingly.
In our example, the function has two parts: one for when the input, called the independent variable, is less than 2, and one for when it is greater than or equal to 2.
In our example, the function has two parts: one for when the input, called the independent variable, is less than 2, and one for when it is greater than or equal to 2.
- If the input value is less than 2, we use the expression \(x^2 - 2\).
- If the input value is 2 or more, we use the expression \(4 + |x - 5|\).
Absolute Value
Absolute value is a concept that measures how far a number is from zero on the number line without considering direction. It's like asking, "What's the distance from zero?" regardless of whether it's left or right.
In the function given, we encounter the absolute value in the expression \(|x - 5|\). The bars \(|...|\) signify absolute value, which results in positive values whether the inner expression is positive or negative.
In the function given, we encounter the absolute value in the expression \(|x - 5|\). The bars \(|...|\) signify absolute value, which results in positive values whether the inner expression is positive or negative.
- If the inside of the absolute value, \((x-5)\), is positive or equal to zero, the absolute value doesn't change it.
- If \((x-5)\) is negative, the value changes to its positive counterpart.
Quadratic Functions
Quadratic functions are polynomial functions of degree 2, and they generally have a standard form of \(ax^2 + bx + c\). They are prominent because they create parabolas when graphed, with the specific shape depending on the coefficients \(a\), \(b\), and \(c\).
In a piecewise function scenario, the part defined by \(f(x) = x^2 - 2\) is a quadratic function. Here, the term \(x^2\) indicates it has a degree of 2, and it's subtracted by a constant, 2.
Special characteristics of quadratic functions include:
In a piecewise function scenario, the part defined by \(f(x) = x^2 - 2\) is a quadratic function. Here, the term \(x^2\) indicates it has a degree of 2, and it's subtracted by a constant, 2.
Special characteristics of quadratic functions include:
- The highest point or lowest point on their curve (depending on the direction of the parabola) is called the vertex.
- If the parabola opens upwards, the vertex is the minimum point; if it opens downwards, the vertex is the maximum.
Other exercises in this chapter
Problem 50
For the following exercises, determine whether the function is odd, even, or neither. $$ f(x)=(x-2)^{2} $$
View solution Problem 50
Determine whether the function is odd, even, or neither. $$f(x)=(x-2)^{2}$$
View solution Problem 51
For the following exercises, determine whether the function is odd, even, or neither. $$ g(x)=2 x^{4} $$
View solution Problem 51
Determine whether the function is odd, even, or neither. $$g(x)=2 x^{4}$$
View solution