Problem 70 - Algebra and Trigonometry

Question

Compositions of Odd and Even Functions Suppose that $$ h=f \circ g $$ If \(g\) is an even function, is \(h\) necessarily even? If \(g\) is odd, is \(h\) odd? What if \(g\) is odd and \(f\) is odd? What if \(g\) is odd and \(f\) is even?

Step-by-Step Solution

Answer

\(h\) is not necessarily even or odd if \(g\) is just even or odd. If both \(g\) and \(f\) are odd, \(h\) is odd; if \(g\) is odd and \(f\) is even, \(h\) is even.

Step 1: Understanding the Composition Function

A composition of functions, denoted as \( h = f \circ g \), means that \( h(x) = f(g(x)) \). This means you first apply the function \( g \) to \( x \), and then apply the function \( f \) to the result of \( g(x) \).

Step 2: Determine if \( h \) is Even when \( g \) is Even

An even function satisfies \( g(-x) = g(x) \). Since \( h(x) = f(g(x)) \), if \( g \) is even, then \( h(-x) = f(g(-x)) = f(g(x)) \). Whether \( h \) is even depends on the function \( f \). Thus, \( h \) is not necessarily even just because \( g \) is even.

Step 3: Determine if \( h \) is Odd when \( g \) is Odd

An odd function satisfies \( g(-x) = -g(x) \). For \( h(x) = f(g(x)) \), if \( g \) is odd, then \( h(-x) = f(g(-x)) = f(-g(x)) \). Whether \( h \) is odd depends on the function \( f \) needs to satisfy \( f(-y) = -f(y) \) for \( y = g(x) \). Thus, \( h \) is not necessarily odd just because \( g \) is odd.

Step 4: Check if \( h \) is Odd with Both \( g \) and \( f \) Odd

If both \( g \) and \( f \) are odd functions, we substitute to get \( h(-x) = f(g(-x)) = f(-g(x)) = -f(g(x)) = -h(x) \). Therefore, \( h(x) \) is odd in this case.

Step 5: Evaluate if \( h \) is Even if \( g \) is Odd and \( f \) is Even

If \( g \) is odd and \( f \) is even, then \( h(-x) = f(g(-x)) = f(-g(x)) = f(g(x)) = h(x) \), so \( h(x) \) is even in this case.

Key Concepts

Even Function

An even function is one that remains unchanged when its input value is negated. In other words, for a function \( g(x) \) to be considered even, it must satisfy the property \( g(-x) = g(x) \) for all elements in its domain. This property basically means that the function has a kind of symmetry with respect to the y-axis, such as a parabola opening upwards.
When we deal with compositions, like when \( h = f \circ g \), understanding whether \( h \) remains even depends on which function in the composition is even or odd. If \( g \) is even, \( g(-x) = g(x) \) doesn't guarantee \( h = f \circ g \) is even unless \( f \) itself holds specific symmetrical properties. Just because \( g(-x) = g(x) \), it doesn't automatically make \( h(x) \) symmetric without further conditions on \( f \).
In particular, for \( h(x) \) to be even in this setup, \( f(g(-x)) \) would also need to equal \( f(g(x)) \). This only happens if \( f \) is constant or otherwise handled symmetrically around \( g(x) \). Therefore, just having \( g \) as even doesn’t necessarily ensure \( h \) will be even.

Odd Function

Odd functions possess a distinct type of symmetry. For a function \( g(x) \) to be considered odd, it must satisfy \( g(-x) = -g(x) \). This means that the function has rotational symmetry around the origin on a graph, like a cubic function that twists around 180 degrees.
In the context of function composition, where \( h(x) = f(g(x)) \), our interest might be to find if \( h \) is odd, given that \( g \) is odd. When \( g \) is odd, this property \( g(-x) = -g(x) \) influences the composition, but doesn't seal the deal for \( h \) being odd without constraints on \( f \). For \( h \) to be odd, \( f(g(-x)) \) should equal \( -f(g(x)) \). Therefore, \( f \) itself needs to be odd to keep the property status consistent.
Notably, if both functions \( f \) and \( g \) involved in the composition are odd, then \( h \) will indeed be odd because this allows the composition to satisfy \( h(-x) = -h(x) \). This wonderful mirroring effect of both odd functions results in \( h \) also being odd, confirming symmetry about the origin across the composition.

Function Properties

The composition of functions involves combining two functions to create a third one. In the composition \( h = f \circ g \), the order is crucial and it means \( h(x) = f(g(x)) \). This functional relationship builds layers onto the output based on the input, providing powerful reconstruction of data or manipulation for desired properties like symmetry.
Function properties such as evenness or oddness can drastically change when functions are composed. A key aspect to consider in compositions is whether the properties like being even or odd are preserved or altered due to the conditions of \( f \) and \( g \).

• **Symmetry Influences**: Even functions provide y-axis symmetry and odd functions offer origin symmetry.
• **Composition Outcomes**: An even \( g \) doesn’t ensure \( h \) is even unless \( f \) complies, and the same scenario for \( g \) odd requires \( f \) odd.
• **Combined Effects**: If both contributing functions are odd, or if one is even and can neutralize oddness through conditions laid upon \( g \), then properties can align in expected ways.

Knowing these underlying function properties helps us predict and understand complex mathematical behavior during composition. With attention to these, we can better anticipate the form and nature of function compositions.