Problem 16 - Algebra 2 and Trigonometry

Question

In \(11-18 :\) a. Find \(h(x)\) when \(h(x)=g(f(x)) .\) b. What is the domain of \(h(x) ?\) c. What is the range of \(\mathrm{h}(x) ?\) d. Graph \(\mathrm{h}(x)\) $$ \mathrm{f}(x)=|2+x|, \mathrm{g}(x)=-x $$

Step-by-Step Solution

Answer

h(x) = -|2+x| has domain (-∞, ∞) and range (-∞, 0].

Step 1: Understand Functions f(x) and g(x)

We have two functions: \(f(x) = |2 + x|\) and \(g(x) = -x\). The function \(f(x)\) represents the absolute value function, which outputs the distance of \(2 + x\) from zero. The function \(g(x)\) takes any input \(x\) and outputs its negation.

Step 2: Compose h(x) = g(f(x))

To find \(h(x)\), substitute \(f(x)\) into \(g(x)\). Since \(f(x) = |2 + x|\), we compose as follows: \(h(x) = g(f(x)) = g(|2 + x|) = -|2 + x|\). So, \(h(x) = -|2+x|\).

Step 3: Find the Domain of h(x)

The domain of \(h(x)\) is determined by the domain of \(f(x)\) since \(g(x)\) operates on the output of \(f(x)\). The function \(f(x) = |2+x|\) is defined for all \(x\), meaning \(h(x)\) is also defined for all \(x\). Therefore, the domain of \(h(x)\) is \((-, )\).

Step 4: Determine the Range of h(x)

The range of \(f(x) = |2+x|\) is \([0, )\) because absolute values are always non-negative. Since \(g(x) = -x\), it negates any value from \(f(x)\). Thus, the range of \(h(x) = -|2+x|\) becomes \((-, 0]\) because negating non-negative values results in non-positive values.

Step 5: Graph h(x)

To graph \(h(x) = -|2 + x|\), first consider the graph of \(y = |2 + x|\), which is a V-shaped graph with vertex at \((-2, 0)\). The graph of \(-|2 + x|\) reflects this graph over the x-axis, making it an inverted V with the same vertex. The graph extends negatively from the y-axis, indicating that all values of \(h(x)\) are non-positive.

Key Concepts

Absolute Value Function

The absolute value function is an essential concept in mathematics that measures the distance a number is from zero on the number line.
It is always non-negative. For any real number or expression inside the absolute value bars, this function outputs its magnitude without considering its sign.
In the case of our function, we have \( f(x) = |2 + x| \).

• The expression \( 2 + x \) simply shifts the input by +2 on the x-axis.
• The result of \( f(x) \) is always zero or larger because absolute values ensure the output is non-negative.

This particular function graphically represents a V-shaped graph centered at the point \(-2\). So, whenever you see absolute value, think about "distance from zero," which must be either zero or positive.

Domain and Range

Understanding domain and range is crucial when dealing with functions like \( h(x) = -|2 + x| \).

• The domain of a function is the set of all possible input values (usually \(x\)) for which the function is defined.
• The range is the set of all possible output values produced by the function.

For our function, the domain of \( f(x) = |2 + x| \) is all real numbers because you can add any real number to 2 and then take its absolute value.
Since \( h(x) = g(f(x)) = -|2 + x| \), the function is also defined for all real numbers, so the domain of \( h(x) \) is \((-\infty, \infty)\).

Range Explained

The range of \( f(x) = |2+x| \) is \([0, \infty)\) because absolute values cannot be negative.
However, the function \( g(x) = -x \) then negates this output, thus transforming the range of \( h(x) \) to \((-\infty, 0]\), reversing the positive values to negative.

Graphing Transformations

Graphing transformations involve shifting, reflecting, or changing the scale of a graph based on mathematical operations applied to the function.
In this exercise, \( h(x) = -|2 + x| \) can be seen as a series of transformations applied to the basic absolute value graph \( y = |x| \).

• The term \( +2 \) inside the absolute bars translates the graph 2 units to the left. This means the point where the graph's two lines meet, or the vertex, moves from the origin \((0,0)\) to \((-2, 0)\).
• The negation sign in front of the absolute value, from \( -|2+x| \), reflects the graph upside down across the x-axis. This reflects the V-shape downwards.

What results is an inverted V-shape graph centered at \((-2,0)\). All the values of \( h(x) \) will be less than or equal to zero, as the graph resides entirely in the non-positive y-area.
Understanding these transformations helps to easily predict the shape and the positions of function graphs.