Is $sqrt{x}$ an even function
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I'm going through some pre-calculus and I am presented with this rule.
If $f(x)$ is an even function, then $f(x) = f(-x)$
So with the following example I have:
$$f(x) = 3x^4 \
f(-x) = 3(-x)^4 = 3x^4$$
However, when I look at the graph of $f(x) = sqrt{-x}$ it is the reflection of $f(x) = sqrt{x}$ about the y axis, but I cannot get this to work out algebraically...
$$f(x) = sqrt{x} = f(-x) = sqrt{-x}$$ which is is true, because of domain restrictions, but $f(x)$ does not look like $f(-x)$ like in the previous example.
Is it because the domain changes for $sqrt{-x}$ that I can do some trick to make turn it into $sqrt{x}$ to allow $f(x) = f(-x)$?
For instance,
But $Dom(f(-x)) = (-infty, 0]$, therefore $-x = x$??
I'm not sure how to finish this off algebraically?
Help anyone?
algebra-precalculus inverse-function
asked Nov 15 at 3:02
Bucephalus
650417
add a comment |
up vote
1
down vote
favorite
I'm going through some pre-calculus and I am presented with this rule.
If $f(x)$ is an even function, then $f(x) = f(-x)$
So with the following example I have:
$$f(x) = 3x^4 \
f(-x) = 3(-x)^4 = 3x^4$$
However, when I look at the graph of $f(x) = sqrt{-x}$ it is the reflection of $f(x) = sqrt{x}$ about the y axis, but I cannot get this to work out algebraically...
$$f(x) = sqrt{x} = f(-x) = sqrt{-x}$$ which is is true, because of domain restrictions, but $f(x)$ does not look like $f(-x)$ like in the previous example.
Is it because the domain changes for $sqrt{-x}$ that I can do some trick to make turn it into $sqrt{x}$ to allow $f(x) = f(-x)$?
For instance,
But $Dom(f(-x)) = (-infty, 0]$, therefore $-x = x$??
I'm not sure how to finish this off algebraically?
Help anyone?
algebra-precalculus inverse-function
asked Nov 15 at 3:02
Bucephalus
650417
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm going through some pre-calculus and I am presented with this rule.
If $f(x)$ is an even function, then $f(x) = f(-x)$
So with the following example I have:
$$f(x) = 3x^4 \
f(-x) = 3(-x)^4 = 3x^4$$
However, when I look at the graph of $f(x) = sqrt{-x}$ it is the reflection of $f(x) = sqrt{x}$ about the y axis, but I cannot get this to work out algebraically...
$$f(x) = sqrt{x} = f(-x) = sqrt{-x}$$ which is is true, because of domain restrictions, but $f(x)$ does not look like $f(-x)$ like in the previous example.
Is it because the domain changes for $sqrt{-x}$ that I can do some trick to make turn it into $sqrt{x}$ to allow $f(x) = f(-x)$?
For instance,
But $Dom(f(-x)) = (-infty, 0]$, therefore $-x = x$??
I'm not sure how to finish this off algebraically?
Help anyone?
algebra-precalculus inverse-function
asked Nov 15 at 3:02
Bucephalus
650417
I'm going through some pre-calculus and I am presented with this rule.
If $f(x)$ is an even function, then $f(x) = f(-x)$
So with the following example I have:
$$f(x) = 3x^4 \
f(-x) = 3(-x)^4 = 3x^4$$
However, when I look at the graph of $f(x) = sqrt{-x}$ it is the reflection of $f(x) = sqrt{x}$ about the y axis, but I cannot get this to work out algebraically...
$$f(x) = sqrt{x} = f(-x) = sqrt{-x}$$ which is is true, because of domain restrictions, but $f(x)$ does not look like $f(-x)$ like in the previous example.
Is it because the domain changes for $sqrt{-x}$ that I can do some trick to make turn it into $sqrt{x}$ to allow $f(x) = f(-x)$?
For instance,
But $Dom(f(-x)) = (-infty, 0]$, therefore $-x = x$??
I'm not sure how to finish this off algebraically?
Help anyone?
algebra-precalculus inverse-function
algebra-precalculus inverse-function
asked Nov 15 at 3:02
Bucephalus
650417
asked Nov 15 at 3:02
Bucephalus
650417
asked Nov 15 at 3:02
Bucephalus
650417
asked Nov 15 at 3:02
Bucephalus
650417
asked Nov 15 at 3:02
Bucephalus
650417
650417
add a comment |
add a comment |
2 Answers
2
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oldest
votes
up vote
5
down vote
accepted
If $f(x)$ is a even function in $mathbb{R}$, as you noticed $f(x)=f(-x)$. Let $x_0$ be a positive number, according to the definition, $f(-x_0)=f(x_0)$ is defined. But clearly $f(-x_0)$ is not defined as $-x_0<0$. So $f$ is not a even function.
answered Nov 15 at 3:06
Kemono Chen
1,637330
Oh yeah of course. I get it now, I was comparing two different functions and not simply looking in the negative domain the the same function. I get it now. Thankyou. @KemonoChen
– Bucephalus
Nov 15 at 3:11
add a comment |
up vote
2
down vote
In a more straightforward manner, for $f$ to be even we need $f(-x)=f(x)$. But
$$f(-x)=sqrt{-x}=pmsqrt{x}i=if(x)$$
and $pm ineq 1$, so clearly it's not even.
answered Nov 15 at 5:00
YiFan
1,5831314
There are two values of $sqrt{-1}$.
– Mark Viola
Nov 15 at 5:29
@MarkViola: $-i ≠1$ either, so the argument still applies.
– Dan
Nov 15 at 5:37
@Dan Yes, I know Dan. That does not excuse an incomplete solution. One should have explained that on one branch of the square root function, $sqrt{-1}=i$
– Mark Viola
Nov 15 at 5:42
You're right, I've edited to reflect this.
– YiFan
Nov 15 at 5:53
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
If $f(x)$ is a even function in $mathbb{R}$, as you noticed $f(x)=f(-x)$. Let $x_0$ be a positive number, according to the definition, $f(-x_0)=f(x_0)$ is defined. But clearly $f(-x_0)$ is not defined as $-x_0<0$. So $f$ is not a even function.
answered Nov 15 at 3:06
Kemono Chen
1,637330
Oh yeah of course. I get it now, I was comparing two different functions and not simply looking in the negative domain the the same function. I get it now. Thankyou. @KemonoChen
– Bucephalus
Nov 15 at 3:11
add a comment |
up vote
5
down vote
accepted
If $f(x)$ is a even function in $mathbb{R}$, as you noticed $f(x)=f(-x)$. Let $x_0$ be a positive number, according to the definition, $f(-x_0)=f(x_0)$ is defined. But clearly $f(-x_0)$ is not defined as $-x_0<0$. So $f$ is not a even function.
answered Nov 15 at 3:06
Kemono Chen
1,637330
Oh yeah of course. I get it now, I was comparing two different functions and not simply looking in the negative domain the the same function. I get it now. Thankyou. @KemonoChen
– Bucephalus
Nov 15 at 3:11
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
If $f(x)$ is a even function in $mathbb{R}$, as you noticed $f(x)=f(-x)$. Let $x_0$ be a positive number, according to the definition, $f(-x_0)=f(x_0)$ is defined. But clearly $f(-x_0)$ is not defined as $-x_0<0$. So $f$ is not a even function.
answered Nov 15 at 3:06
Kemono Chen
1,637330
If $f(x)$ is a even function in $mathbb{R}$, as you noticed $f(x)=f(-x)$. Let $x_0$ be a positive number, according to the definition, $f(-x_0)=f(x_0)$ is defined. But clearly $f(-x_0)$ is not defined as $-x_0<0$. So $f$ is not a even function.
answered Nov 15 at 3:06
Kemono Chen
1,637330
answered Nov 15 at 3:06
Kemono Chen
1,637330
answered Nov 15 at 3:06
Kemono Chen
1,637330
answered Nov 15 at 3:06
Kemono Chen
1,637330
1,637330
Oh yeah of course. I get it now, I was comparing two different functions and not simply looking in the negative domain the the same function. I get it now. Thankyou. @KemonoChen
– Bucephalus
Nov 15 at 3:11
add a comment |
Oh yeah of course. I get it now, I was comparing two different functions and not simply looking in the negative domain the the same function. I get it now. Thankyou. @KemonoChen
– Bucephalus
Nov 15 at 3:11
Oh yeah of course. I get it now, I was comparing two different functions and not simply looking in the negative domain the the same function. I get it now. Thankyou. @KemonoChen
– Bucephalus
Nov 15 at 3:11
Oh yeah of course. I get it now, I was comparing two different functions and not simply looking in the negative domain the the same function. I get it now. Thankyou. @KemonoChen
– Bucephalus
Nov 15 at 3:11
add a comment |
up vote
2
down vote
In a more straightforward manner, for $f$ to be even we need $f(-x)=f(x)$. But
$$f(-x)=sqrt{-x}=pmsqrt{x}i=if(x)$$
and $pm ineq 1$, so clearly it's not even.
answered Nov 15 at 5:00
YiFan
1,5831314
There are two values of $sqrt{-1}$.
– Mark Viola
Nov 15 at 5:29
@MarkViola: $-i ≠1$ either, so the argument still applies.
– Dan
Nov 15 at 5:37
@Dan Yes, I know Dan. That does not excuse an incomplete solution. One should have explained that on one branch of the square root function, $sqrt{-1}=i$
– Mark Viola
Nov 15 at 5:42
You're right, I've edited to reflect this.
– YiFan
Nov 15 at 5:53
add a comment |
up vote
2
down vote
In a more straightforward manner, for $f$ to be even we need $f(-x)=f(x)$. But
$$f(-x)=sqrt{-x}=pmsqrt{x}i=if(x)$$
and $pm ineq 1$, so clearly it's not even.
answered Nov 15 at 5:00
YiFan
1,5831314
There are two values of $sqrt{-1}$.
– Mark Viola
Nov 15 at 5:29
@MarkViola: $-i ≠1$ either, so the argument still applies.
– Dan
Nov 15 at 5:37
@Dan Yes, I know Dan. That does not excuse an incomplete solution. One should have explained that on one branch of the square root function, $sqrt{-1}=i$
– Mark Viola
Nov 15 at 5:42
You're right, I've edited to reflect this.
– YiFan
Nov 15 at 5:53
add a comment |
up vote
2
down vote
up vote
2
down vote
In a more straightforward manner, for $f$ to be even we need $f(-x)=f(x)$. But
$$f(-x)=sqrt{-x}=pmsqrt{x}i=if(x)$$
and $pm ineq 1$, so clearly it's not even.
answered Nov 15 at 5:00
YiFan
1,5831314
In a more straightforward manner, for $f$ to be even we need $f(-x)=f(x)$. But
$$f(-x)=sqrt{-x}=pmsqrt{x}i=if(x)$$
and $pm ineq 1$, so clearly it's not even.
answered Nov 15 at 5:00
YiFan
1,5831314
edited Nov 15 at 5:52
answered Nov 15 at 5:00
YiFan
1,5831314
answered Nov 15 at 5:00
YiFan
1,5831314
answered Nov 15 at 5:00
YiFan
1,5831314
1,5831314
There are two values of $sqrt{-1}$.
– Mark Viola
Nov 15 at 5:29
@MarkViola: $-i ≠1$ either, so the argument still applies.
– Dan
Nov 15 at 5:37
@Dan Yes, I know Dan. That does not excuse an incomplete solution. One should have explained that on one branch of the square root function, $sqrt{-1}=i$
– Mark Viola
Nov 15 at 5:42
You're right, I've edited to reflect this.
– YiFan
Nov 15 at 5:53
add a comment |
There are two values of $sqrt{-1}$.
– Mark Viola
Nov 15 at 5:29
@MarkViola: $-i ≠1$ either, so the argument still applies.
– Dan
Nov 15 at 5:37
@Dan Yes, I know Dan. That does not excuse an incomplete solution. One should have explained that on one branch of the square root function, $sqrt{-1}=i$
– Mark Viola
Nov 15 at 5:42
You're right, I've edited to reflect this.
– YiFan
Nov 15 at 5:53
There are two values of $sqrt{-1}$.
– Mark Viola
Nov 15 at 5:29
There are two values of $sqrt{-1}$.
– Mark Viola
Nov 15 at 5:29
@MarkViola: $-i ≠1$ either, so the argument still applies.
– Dan
Nov 15 at 5:37
@MarkViola: $-i ≠1$ either, so the argument still applies.
– Dan
Nov 15 at 5:37
@Dan Yes, I know Dan. That does not excuse an incomplete solution. One should have explained that on one branch of the square root function, $sqrt{-1}=i$
– Mark Viola
Nov 15 at 5:42
@Dan Yes, I know Dan. That does not excuse an incomplete solution. One should have explained that on one branch of the square root function, $sqrt{-1}=i$
– Mark Viola
Nov 15 at 5:42
You're right, I've edited to reflect this.
– YiFan
Nov 15 at 5:53
You're right, I've edited to reflect this.
– YiFan
Nov 15 at 5:53
add a comment |
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