Are f(x) and f(-x) Equivalent Functions?

  • Context: High School 
  • Thread starter Thread starter dyn
  • Start date Start date
  • Tags Tags
    Argument Function
Click For Summary
SUMMARY

Functions f(x) and f(-x) are not equivalent; they represent different mathematical entities. For example, if f(x) = x² + x, then f(-x) = x² - x, indicating that they yield different outputs for the same input. However, the graphs of f(x) and f(-x) are mirror images across the y-axis. This property holds true for all functions, including trigonometric functions like sin(x), where f(-x) = -sin(x), demonstrating that they are also not equivalent but exhibit symmetry.

PREREQUISITES
  • Understanding of basic function notation and evaluation
  • Knowledge of even and odd functions
  • Familiarity with graphing functions and their transformations
  • Concept of function composition
NEXT STEPS
  • Study the properties of even and odd functions in detail
  • Learn about function composition and its implications
  • Explore graphical transformations of functions
  • Investigate the symmetry of trigonometric functions
USEFUL FOR

Mathematics students, educators, and anyone interested in understanding function properties and their graphical representations.

dyn
Messages
774
Reaction score
63
Hi.
If I have a function for example f( x ) = x2 + x then to obtain f( -x) I just put (-x) in place of the x in f(x)
so I get f( -x) = x2 - x
Am I right so far ?
So f(x) and f(-x) look like different functions but if you put a negative number in f(-x) it flips the -x back to +x so are f(x) and f(-x) are equivalent ?
But if I put sin(x) as the argument I get a totally different function

Thanks
 
Mathematics news on Phys.org
dyn said:
Hi.
If I have a function for example f( x ) = x2 + x then to obtain f( -x) I just put (-x) in place of the x in f(x)
so I get f( -x) = x2 - x
Am I right so far ?
Yes
So f(x) and f(-x) look like different functions
Yes. In fact, they are different functions. Which suggests that it is not a good idea to call them both by the same name. So invent a different name. Let g(x) = f(-x).

Now f(x) is a different function from g(x). In particular, f(1) = 12 + 1 = 2 and g(1) = f(-1) = (-1)2 + (-1) = 0.
but if you put a negative number in f(-x) it flips the -x back to +x so are f(x) and f(-x) are equivalent ?
Yes, if you put a negative number into g(x) and the corresponding positive number into f(x) you get the same result out. The graphs of the two functions are mirror images of each other, reflected across the y axis.

g(x) = f(-x) and f(-x) = g(x).
But if I put sin(x) as the argument I get a totally different function
You lost me. What function are you getting when you do what to what and how is it different from what?

You may be talking about something called function composition where you apply two functions in sequence to the input argument. So if "-" is the name of the additive inverse function "-"(x) = -x then your original example could be thought of as the composition of f with "-": ##g = f \circ -## or g(x) = f(-x).

And your sin example could be thought of as the composition of f with sin: ##g = f \circ \sin## or ##g(x) = f(\sin x)##
 
Last edited:
  • Like
Likes   Reactions: dyn
dyn said:
but if you put a negative number in f(-x) it flips the -x back to +x so are f(x) and f(-x) are equivalent ?
They are mirror versions of each other: If you mirror f(x) at the vertical axis you get f(-x). This is a general property of all functions (from real numbers to real numbers).
If f(x)=sin(x), then f(-x)=sin(-x)=-sin(x) which is the sine mirrored at the vertical axis.
 
  • Like
Likes   Reactions: dyn

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K