Determine whether the given function is odd or even

[chwala]

Homework Statement

consider the function:

$f(x)=x^3+3x−1$

Relevant Equations

odd or even functions concept

$f(x)=x^3+3x−1$

 

[chwala]

Ok this was a question i saw on one paper...i will post the question and corresponding solution here:

which is fine with me, i can follow the steps.
My question is, ...is the question also analogous in asking whether the function is even or odd? in that case,
can one use $f(-x)=-f(x)$? the function is not an even function because
$f(-x) ≠ f(x)$

on the other hand,
$f(-x)=-x^3-3x-1$
$-f(x)=-x^3-3x+1$=$-(x^3+3x)-1$...=$-f(x)$ looks a bit interesting...

 

[SammyS]

My question is, ...is the question also analogous in asking whether the function is even or odd? in that case,
can one use $f(-x)=-f(x)$? the function is not an even function because
$f(-x)=f(x)$

on the other hand,
$f(-x)=-x^3-3x-1$
$-f(x)=-x^3-3x+1$=-(x^3+3x)-1$...=-f(x)$ looks a bit interesting...

$-f(x) \ne f(-x)$, so $f$ is not an odd function.

That should answer your question. The function is neither an odd nor even function.

 

[chwala]

interesting, good learning point ,noted... thanks Sammy. My last deduction
$-f(x)=-x^3-3x+1$=$-(x^3+3x)-1$=$-f(x)$ was not correct.

 

[Mark44]

If the given function had been $f(x) = x^3 + 3x$, it's easy to show that this is an odd function by use of the definition. Additionally, both $x^3$ and $3x$, taken as functions on their own, are odd functions (i.e., their own reflection across the origin), and their sum is also an odd function.

However, $f(x) = x^3 + x - 1$ is neither odd nor even, as has already been shown. The last term, $-1$, taken on its own, is an even function, and this prevents $f(x) = x^3 + x - 1$ from being its own reflection across the origin and across the y-axis, so it is neither odd nor even.

 

[chwala]

If the given function had been $f(x) = x^3 + 3x$, it's easy to show that this is an odd function by use of the definition. Additionally, both $x^3$ and $3x$, taken as functions on their own, are odd functions (i.e., their own reflection across the origin), and their sum is also an odd function.

However, $f(x) = x^3 + x - 1$ is neither odd nor even, as has already been shown. The last term, $-1$, taken on its own, is an even function, and this prevents $f(x) = x^3 + x - 1$ from being its own reflection across the origin and across the y-axis, so it is neither odd nor even.

Mark just confirm, last term$-1$ is an even function? am not getting this...
It is a constant and not a function...clarify on this.

 

[Delta2]

The constant function $g(x)=-1$ is even because $$g(x)=-1=g(-x)$$.

 

[chwala]

ok, i will take this with a grain of salt, new to me . $g$ is a function of $x$, of which clearly $-1$ is not. I guess maybe i need to check more on this delta.

 

[Delta2]

ok, i will take this with a grain of salt, new to me . $g$ is a function of $x$, of which clearly $-1$ is not. I guess maybe i need to check more on this delta.

ehm any constant number can be considered to be function of any variable, for example -1 can be considered to be a constant function of x, $g(x)=-1$ and also a constant function of z $h(z)=-1$.

 

[SammyS]

ok, i will take this with a grain of salt, new to me . $g$ is a function of $x$, of which clearly $-1$ is not. I guess maybe i need to check more on this delta.

chwala,
Maybe this helps.

Write the constant function, $g(x)=-1$ in slope-intercept form.

$g(x)=0\cdot (x)-1$

 

[chwala]

thanks new to me though, cheers