The given function is an even or an odd function?

AI Thread Summary
The discussion centers on determining whether the function defined as g(x) = f(x^2) is even or odd. Participants agree that since x^2 is an even function, g(x) will also be even, as shown by the relationship g(-x) = g(x). The formal proof involves using the definitions of even and odd functions to demonstrate that f((-x)^2) = f(x^2). The original poster expresses gratitude for the assistance and confirms that they have solved the problem. The conversation concludes with no further replies needed.
brochesspro
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Homework Statement
Determine whether the function ##f(x^2)## is an even function, or an odd function, or neither, where ##f(x)## is an arbitrary function.
Relevant Equations
Given below.
1641373337430.png

I think the answer is an even function as the function ##x^2## is an even function and thus, is symmetrical w.r.t. Y axis. The question I have is how to do this problem algebraically. I tried to graph some functions on GeoGebra to verify my answer.

a) ##y = ln(x^2)##
1641373734325.png


b) ##y = sin(x^2)##
1641373860150.png


These are the two functions I tried, and both look pretty even to me. Please help me solve this problem. Thank you.
 
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How is the function ##f(x^2)## defined?
 
PS how did the computer calculate the function values to draw that graph for you?
 
PeroK said:
PS how did the computer calculate the function values to draw that graph for you?

The computer first calculates the values of the squares of the input value and applies the arbitrary function to the new value. I know that since the function ##x^2## is an even function and thus, is symmetrical w.r.t. Y axis, the output of the arbitrary function will be even. Eg. ##f((-1)^2) = f(1^2)## as ##(-1)^2 = 1^2 = 1##. I am sure I am correct till here. Are you?

I will see you after two hours.
 
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brochesspro said:
The computer first calculates the values of the squares of the input value and applies the arbitrary function to the new value.
Exactly. This is called the composition of two functions. Technically, we define a new function ##g(x)## by:$$g(x) = f(x^2)$$And then ##g(x)## is an even function for any function ##f(x)##, as you have shown for ##x = 1##.
 
PeroK said:
Exactly. This is called the composition of two functions. Technically, we define a new function ##g(x)## by:$$g(x) = f(x^2)$$And then ##g(x)## is an even function for any function ##f(x)##, as you have shown for ##x = 1##.
But how do I write it formally?
 
You should directly use the definitions of "even" and "odd" functions. Start with ##g(-x) = f((-x)^2)## and see where that leads you for the function g. Is ##g(x)## even or odd or neither?
 
brochesspro said:
But how do I write it formally?
##f((-x)^2) = f(x^2)##, for ##x^2## in the domain of f.
Thus ##f(x^2)## is an even function.
 
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Mark44 said:
##f((-x)^2) = f(x^2)##, for ##x^2## in the domain of f.
Thus ##f(x^2)## is an even function.
FactChecker said:
You should directly use the definitions of "even" and "odd" functions. Start with ##g(-x) = f((-x)^2)## and see where that leads you for the function g. Is ##g(x)## even or odd or neither?
Oh, I see. Thank you.
 
  • #10
To do it fully formally you have first to define the function $$g(x)=f(x^2)$$ and then prove that $$g(-x)=g(x)$$ and hence g is even. The proof is half a line long, it is essentially what @Mark44 does at post #8.
 
  • #11
Delta2 said:
To do it fully formally you have first to define the function $$g(x)=f(x^2)$$ and then prove that $$g(-x)=g(x)$$ and hence g is even. The proof is half a line long, it is essentially what @Mark44 does at post #8.
And precisely what I did in post #5.
 
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  • #12
Thank you for your replies. They mean a lot to me. I solved the problem long ago, so no need to reply to this thread any further.
 
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