If f and g are functions defined for all real numbers, and f is an odd function,
then f ∘ g is also an odd function. Justify.
I wrote false cause for example if f = x^3(odd) and g=x^2(even) the fog is even.
you are correct.
as an even simpler example, let
g(x) = x-1 (neither even nor odd)
f(x) = x (odd)
f∘g = x-1
Cheers Steve.
To decide whether the statement "f ∘ g is an odd function" is true or false, we need to understand the concept of function composition and consider the properties of odd functions.
Function composition is a way to combine two functions, where the output of one function is used as the input for the other. In this case, we are interested in the composition of f and g, which is denoted as f ∘ g (read as "f composed with g").
An odd function is defined as a function that satisfies the property f(-x) = -f(x) for all x in its domain. In other words, if we replace x with its negative, the function's output will be the negative of its original output.
To determine whether f ∘ g is an odd function, we can analyze its property:
(f ∘ g)(-x) = f(g(-x))
Since g is defined for all real numbers, we can evaluate g(-x) by replacing x with -x in g. Therefore, g(-x) = g(x).
Now let's substitute this result into the expression for (f ∘ g)(-x):
(f ∘ g)(-x) = f(g(-x)) = f(g(x))
At this point, we realize that the composition f ∘ g does not have the same property as an odd function. In general, f ∘ g will not be odd even if f is odd.
To illustrate this further, your example of f(x) = x^3 (an odd function) and g(x) = x^2 (an even function) is an excellent demonstration.
(f ∘ g)(x) = f(g(x)) = f(x^2) = (x^2)^3 = x^6
By evaluating (f ∘ g)(-x), we can see that:
(f ∘ g)(-x) = (-x)^6 = x^6
Here, the outputs are equivalent, which means (f ∘ g) does not satisfy the property of an odd function.
Therefore, you are correct that the statement "f ∘ g is an odd function" is false. The composition of an odd function and any function, including an even function, does not guarantee that the resulting function will be odd.