Question

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?

Short answer

The function h is even function when g is an even function.

The function h cannot be determined when g is odd function.

The function h is odd function when both f and g are odd function.

The function h is even function when the function f is even function and function g is odd function.

Step-by-step solution

Step 1. Apply the concept of even and odd functions.

A function is said to be even function when$f\left(-x\right)=f\left(x\right)$ that is when x is replaced by$-x$ the function does not change.

A function is said to be odd function when$f\left(-x\right)=-f\left(x\right)$ that is when x is replaced by$-x$ the function becomes negative of original function.

Step 2. Apply the concept of composite function.

The composition of two functions g and fis expressed as $g\circ f$. This is also called composite function. Mathematically it is expressed below.

width="127">g∘fx=gfx

Step 3. Interpret the nature of function.

Consider the composition of the function provided $f\circ g=h$.

This is rewritten as $\left(f\circ g\right)\left(x\right)=h\left(x\right)$.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\end{array}$

Now, it is provided that the function g is an even function so $g\left(-x\right)=g\left(x\right)$.

Now, interpret the nature of function h.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\end{array}$

Replace xby $-x$.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\\ f\left(g\left(-x\right)\right)=h\left(-x\right)\end{array}$

Since, g is an even function so $g\left(-x\right)=g\left(x\right)$.

$\begin{array}{c}f\left(g\left(-x\right)\right)=h\left(-x\right)\\ f\left(g\left(x\right)\right)=h\left(-x\right)\end{array}$

Replace left hand side by $h\left(x\right)$.

$h\left(x\right)=h\left(-x\right)$

As,$h\left(x\right)=h\left(-x\right)$ and this is the property of even function so h is an even function.

Thus, the function h is even function when g is an even function.

Step 4. Interpret the nature of function when g is odd.

Consider the composition of the function provided $f\circ g=h$.

This is rewritten as $\left(f\circ g\right)\left(x\right)=h\left(x\right)$.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\end{array}$

Now, it is provided that the function g is an odd function so $g\left(-x\right)=-g\left(x\right)$.

Now, interpret the nature of function h.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\end{array}$

Replace xby $-x$.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\\ f\left(g\left(-x\right)\right)=h\left(-x\right)\end{array}$

Since, g is an odd function so $g\left(-x\right)=-g\left(x\right)$.

$\begin{array}{c}f\left(g\left(-x\right)\right)=h\left(-x\right)\\ f\left(-g\left(x\right)\right)=h\left(-x\right)\end{array}$

This cannot be further simplified as it is not defined where function f is even or odd.

Thus, the function h cannot be determined when g is odd function.

Step 5. Interpret the nature of function when g is odd and f is odd.

Consider the composition of the function provided $f\circ g=h$.

This is rewritten as $\left(f\circ g\right)\left(x\right)=h\left(x\right)$.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\end{array}$

Now, it is provided that the function f and g are odd functions so$g\left(-x\right)=-g\left(x\right)$ and$f\left(-x\right)=-f\left(x\right)$

Now, interpret the nature of function h.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\end{array}$

Replace xby $-x$.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\\ f\left(g\left(-x\right)\right)=h\left(-x\right)\end{array}$

Since, g is an odd function so $g\left(-x\right)=-g\left(x\right)$.

$\begin{array}{c}f\left(g\left(-x\right)\right)=h\left(-x\right)\\ f\left(-g\left(x\right)\right)=h\left(-x\right)\end{array}$

Since, f is an odd function so $f\left(-x\right)=-f\left(x\right)$.

$\begin{array}{c}f\left(g\left(-x\right)\right)=h\left(-x\right)\\ f\left(-g\left(x\right)\right)=h\left(-x\right)\\ -f\left(g\left(x\right)\right)=h\left(-x\right)\end{array}$

Replace left hand side by $h\left(x\right)$.

$-h\left(x\right)=h\left(-x\right)$

As,$-h\left(x\right)=h\left(-x\right)$ and this is the property of odd function so h is an odd function.

Thus, the function h is odd function when both f and g are odd function.

Step 6. Interpret the nature of function when g is odd and f is even.

Consider the composition of the function provided $f\circ g=h$.

This is rewritten as $\left(f\circ g\right)\left(x\right)=h\left(x\right)$.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\end{array}$

Now, it is provided that the function f is even and g is odd function so$g\left(-x\right)=-g\left(x\right)$ and$f\left(-x\right)=f\left(x\right)$

Now, interpret the nature of function h.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\end{array}$

Replace xby $-x$.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=h\left(x\right)\\ f\left(g\left(x\right)\right)=h\left(x\right)\\ f\left(g\left(-x\right)\right)=h\left(-x\right)\end{array}$

Since, g is an odd function so $g\left(-x\right)=-g\left(x\right)$.

$\begin{array}{c}f\left(g\left(-x\right)\right)=h\left(-x\right)\\ f\left(-g\left(x\right)\right)=h\left(-x\right)\end{array}$

Since, f is an even function so $f\left(-x\right)=f\left(x\right)$.

$\begin{array}{c}f\left(g\left(-x\right)\right)=h\left(-x\right)\\ f\left(-g\left(x\right)\right)=h\left(-x\right)\\ f\left(g\left(x\right)\right)=h\left(-x\right)\end{array}$

Replace left hand side by $h\left(x\right)$.

$h\left(x\right)=h\left(-x\right)$

As,$h\left(x\right)=h\left(-x\right)$ and this is the property of even function so h is an even function.

Thus, the function h is even function when the function f is even function and function g is odd function.