Problem 44 - Contemporary Precalculus

Question

$$\text { If } g(x)=\left\\{\begin{array}{ll} 2 x-3 & \text { if } x<-1 \\ |x|-5 & \text { if }-1 \leq x \leq 2 \\ x^{2} & \text { if } x>2 \end{array} \quad\text { find }\right.$$ (a) The domain of \(g\) (b) \(g(-2.5)\) (c) \(g(-1)\) (d) \(g(2)\) (e) \(g(4)\)

Step-by-Step Solution

Answer

(a) Domain of \(g\): __________________ (b) \(g(-2.5) =\) __________________ (c) \(g(-1) =\) __________________ (d) \(g(2) =\) __________________ (e) \(g(4) =\) __________________

Step 1: Determine the domain of \(g\)

The domain of \(g\) is the set of all possible input values for which the function is defined. In this case, the function is defined for all real numbers, since each case of the piecewise function covers a different interval of the real line. Therefore, the domain of \(g\) is: $$\text{Domain }(g) = \{-\infty < x < \infty\}.$$

Step 2: Evaluate \(g(-2.5)\)

To find \(g(-2.5)\), we need to determine which case of the piecewise function to use for \(x=-2.5\). Since \(x\) falls in the interval \(x <-1\), we use the first case (where the function is defined as \(2x - 3\)): $$g(-2.5) = 2(-2.5) - 3 = -5 - 3 = -8.$$

Step 3: Evaluate \(g(-1)\)

To find \(g(-1)\), we use the second case of the piecewise function, since \(-1\) falls in the interval \(-1 \leq x \leq 2\). For this interval, the function is defined as \(|x| - 5\): $$g(-1) = |-1| - 5 = 1 - 5 = -4.$$

Step 4: Evaluate \(g(2)\)

Again, to find \(g(2)\), we use the second case of the piecewise function, since \(2\) falls in the interval \(-1 \leq x \leq 2\). For this interval, the function is defined as \(|x| - 5\): $$g(2) = |2| - 5 = 2 - 5 = -3.$$

Step 5: Evaluate \(g(4)\)

To find \(g(4)\), we use the third case of the piecewise function, since \(4\) falls in the interval \(x>2\). For this interval, the function is defined as \(x^2\): $$g(4) = 4^2 = 16.$$ So, the answers are: (a) The domain of \(g\) is \(\{-\infty < x < \infty\}\). (b) \(g(-2.5) = -8\). (c) \(g(-1) = -4\). (d) \(g(2) = -3\). (e) \(g(4) = 16\).

Key Concepts

Domain of a Function

The domain of a function refers to the complete set of possible values of the independent variable, which is often denoted as \( x \). In simpler terms, it's all about understanding "what can \( x \) be?"
For functions, especially piecewise ones like \( g(x) \), defining the domain means looking at all the ranges where different rules apply. Each part of the piecewise function corresponds to a specific interval.
In the example for \( g(x) \), the function is defined:

• For \( x < -1 \), it uses \( 2x - 3 \).
• For \(-1 \leq x \leq 2\), it uses \( |x|-5 \).
• For \( x > 2 \), it uses \( x^2 \).

Because \( g(x) \) covers all these intervals without missing any parts on the real number line, the domain of \( g \) is all real numbers. So, we can say the domain is \[-\infty < x < \infty.\]
It is important to check each segment individually, ensuring they connect to cover the entire real line.

Absolute Value Function

The absolute value function is one of the key elements in the piecewise function we’re examining. It is symbolized by \( |x| \). The absolute value of a number is its distance from zero on the number line, without considering direction. Therefore, it always results in a non-negative value.
For example, both \( |5| \) and \( |-5| \) are 5. It shows how we remove any negative signs while considering absolute distance. In our piecewise function, for the interval \(-1 \leq x \leq 2\), the function is written as \( |x| - 5 \).
Evaluating this part of the function means first finding \(|x|\) and then subtracting 5.
It’s noteworthy how the absolute value causes a bend or point in graphs, providing a characteristic "V" shape when graphed alone.
As a student, make sure to correctly address the shift caused by subtracting or adding values to the absolute value to understand the transformations of the graph.

Quadratic Function

A quadratic function is typically expressed in the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). It is characterized by its highest degree being 2, hence "quadratic", meaning square. In our exercise, for the interval \( x > 2 \), the function used is \( x^2 \).
Quadratic functions generate a parabola when graphed. This parabola is symmetric about its vertex, and its shape (opening up or down) is determined by whether \( a \) is positive or negative.
Due to the simple form here (just \( x^2 \)), the vertex is at the origin (0,0), and as \( x \) increases or decreases, \( x^2 \) increases, indicating the parabola opens upwards.
Quadratic functions are critical in mathematics as they model various real-world scenarios like projectile motion or calculating areas.

Function Evaluation

Function evaluation involves substituting a given input into a function to determine the output (or value) the function provides. Each piecewise segment of \( g(x) \) requires us to follow specific rules based on \( x \)'s value to correctly evaluate it.
Let’s look at a few examples learnt from the exercise:

• For \( g(-2.5) \), since \( -2.5 < -1 \), we use the first form \( 2x - 3 \), giving us \( g(-2.5) = -8 \).
• For \( g(-1) \), as \( -1 \leq x \leq 2 \), use \( |x| - 5 \), resulting in \( g(-1) = -4 \).
• Finally, for \( g(4) \), because \( 4 > 2 \), use \( x^2 \), which yields \( g(4) = 16 \).

Each calculation thoroughly follows the respective piece of the function. By setting \( x \) to the specified value and applying the appropriate rule, function evaluation can effectively resolve the function’s output.