Proof $y=x^3/(x^2+1)$ is injective
$begingroup$
i need to proof that $$f(x)=frac{x^3}{x^2+1}$$ is injective
I tried $f(x_1)=f(x_2)$ and I also tried with Rolle's theorem.
functions
$endgroup$
add a comment |
$begingroup$
i need to proof that $$f(x)=frac{x^3}{x^2+1}$$ is injective
I tried $f(x_1)=f(x_2)$ and I also tried with Rolle's theorem.
functions
$endgroup$
add a comment |
$begingroup$
i need to proof that $$f(x)=frac{x^3}{x^2+1}$$ is injective
I tried $f(x_1)=f(x_2)$ and I also tried with Rolle's theorem.
functions
$endgroup$
i need to proof that $$f(x)=frac{x^3}{x^2+1}$$ is injective
I tried $f(x_1)=f(x_2)$ and I also tried with Rolle's theorem.
functions
functions
edited Dec 16 '18 at 14:05
egreg
184k1486205
184k1486205
asked Dec 16 '18 at 14:00
shayshay
103
103
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Assume $exists a,b: f(a)=f(b)$, then begin{align*}frac{a^3}{a^2+1}=frac{b^3}{b^2+1}&iff a^3b^2+a^3=b^3a^2+b^3 \ &iff a^3b^2-a^2b^3+a^3-b^3=0 \&iff a^2b^2(a-b)+(a-b)(a^2-ab+b^2)=0 \ &iff (a-b)left[a^2b^2+a^2+ab+b^2right]=0\&iff a=bvee a^2b^2+a^2+ab+b^2=0end{align*}
Edit: based on @egreg's comment: $$a^2b^2+a^2+ab+b^2=(ab)^2+left(a+frac{b}{2}right)^2+frac{3b^2}{4}ge 0$$ and equality holds if and only if $a=b=0$.
Either way, we found that $a=b$, hence the function is injective.
$endgroup$
2
$begingroup$
$a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
$endgroup$
– egreg
Dec 16 '18 at 14:46
$begingroup$
@egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
$endgroup$
– Galc127
Dec 16 '18 at 14:48
add a comment |
$begingroup$
Hint: Prove that, for each real $x$, $f'(x)geqslant0$.
$endgroup$
$begingroup$
I did it, but still dont know how it's can help me
$endgroup$
– shay
Dec 16 '18 at 14:08
1
$begingroup$
SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
$endgroup$
– José Carlos Santos
Dec 16 '18 at 14:10
add a comment |
$begingroup$
Here is an approach if you want to avoid derivatives. For $x neq 0$, we have
$$frac{x^3}{x^2 + 1} = frac{1}{frac{1}{x} + frac{1}{x^3}}.$$
The function $1/x + 1/x^3$ for $x > 0$ is the sum of two strictly decreasing functions, hence itself strictly decreasing and therefore one to one with range $(0, infty)$. Hence $1/(1/x + 1/x^3)$ is strictly increasing and one to one with range $(0, infty)$. For $x < 0$, notice $1/x + 1/x^3$ is an odd function, and therefore a similar argument shows $1/(1/x + 1/x^3)$ is one to one with range $(-infty, 0)$.
In fact, this shows that if you add back the point $x = 0$, the function is a bijection since the function is zero here.
$endgroup$
add a comment |
$begingroup$
Suppose $f(a)=f(b)$: then
$$
a^3(b^2+1)=b^3(a^2+1)
$$
This implies
$$
a^3b^2-a^2b^3+a^3-b^3=0
$$
and so
$$
a^2b^2(a-b)+(a-b)(a^2+ab+b^2)=0
$$
Conclude.
If you know that $a^2+ab+b^2le0$ if and only if $a=b=0$, you're done.
With calculus,
$$
f'(x)=frac{3x^2(x^2+1)-2x^4}{(x^2+1)^2}
$$
The derivative only vanishes at $0$, so Rolle's theorem would only apply to an interval of the form $[a,b]$, with $a<0$ and $b>0$. Can $f(a)=f(b)$ in this case?
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Assume $exists a,b: f(a)=f(b)$, then begin{align*}frac{a^3}{a^2+1}=frac{b^3}{b^2+1}&iff a^3b^2+a^3=b^3a^2+b^3 \ &iff a^3b^2-a^2b^3+a^3-b^3=0 \&iff a^2b^2(a-b)+(a-b)(a^2-ab+b^2)=0 \ &iff (a-b)left[a^2b^2+a^2+ab+b^2right]=0\&iff a=bvee a^2b^2+a^2+ab+b^2=0end{align*}
Edit: based on @egreg's comment: $$a^2b^2+a^2+ab+b^2=(ab)^2+left(a+frac{b}{2}right)^2+frac{3b^2}{4}ge 0$$ and equality holds if and only if $a=b=0$.
Either way, we found that $a=b$, hence the function is injective.
$endgroup$
2
$begingroup$
$a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
$endgroup$
– egreg
Dec 16 '18 at 14:46
$begingroup$
@egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
$endgroup$
– Galc127
Dec 16 '18 at 14:48
add a comment |
$begingroup$
Assume $exists a,b: f(a)=f(b)$, then begin{align*}frac{a^3}{a^2+1}=frac{b^3}{b^2+1}&iff a^3b^2+a^3=b^3a^2+b^3 \ &iff a^3b^2-a^2b^3+a^3-b^3=0 \&iff a^2b^2(a-b)+(a-b)(a^2-ab+b^2)=0 \ &iff (a-b)left[a^2b^2+a^2+ab+b^2right]=0\&iff a=bvee a^2b^2+a^2+ab+b^2=0end{align*}
Edit: based on @egreg's comment: $$a^2b^2+a^2+ab+b^2=(ab)^2+left(a+frac{b}{2}right)^2+frac{3b^2}{4}ge 0$$ and equality holds if and only if $a=b=0$.
Either way, we found that $a=b$, hence the function is injective.
$endgroup$
2
$begingroup$
$a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
$endgroup$
– egreg
Dec 16 '18 at 14:46
$begingroup$
@egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
$endgroup$
– Galc127
Dec 16 '18 at 14:48
add a comment |
$begingroup$
Assume $exists a,b: f(a)=f(b)$, then begin{align*}frac{a^3}{a^2+1}=frac{b^3}{b^2+1}&iff a^3b^2+a^3=b^3a^2+b^3 \ &iff a^3b^2-a^2b^3+a^3-b^3=0 \&iff a^2b^2(a-b)+(a-b)(a^2-ab+b^2)=0 \ &iff (a-b)left[a^2b^2+a^2+ab+b^2right]=0\&iff a=bvee a^2b^2+a^2+ab+b^2=0end{align*}
Edit: based on @egreg's comment: $$a^2b^2+a^2+ab+b^2=(ab)^2+left(a+frac{b}{2}right)^2+frac{3b^2}{4}ge 0$$ and equality holds if and only if $a=b=0$.
Either way, we found that $a=b$, hence the function is injective.
$endgroup$
Assume $exists a,b: f(a)=f(b)$, then begin{align*}frac{a^3}{a^2+1}=frac{b^3}{b^2+1}&iff a^3b^2+a^3=b^3a^2+b^3 \ &iff a^3b^2-a^2b^3+a^3-b^3=0 \&iff a^2b^2(a-b)+(a-b)(a^2-ab+b^2)=0 \ &iff (a-b)left[a^2b^2+a^2+ab+b^2right]=0\&iff a=bvee a^2b^2+a^2+ab+b^2=0end{align*}
Edit: based on @egreg's comment: $$a^2b^2+a^2+ab+b^2=(ab)^2+left(a+frac{b}{2}right)^2+frac{3b^2}{4}ge 0$$ and equality holds if and only if $a=b=0$.
Either way, we found that $a=b$, hence the function is injective.
edited Dec 16 '18 at 14:50
answered Dec 16 '18 at 14:16
Galc127Galc127
3,3121334
3,3121334
2
$begingroup$
$a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
$endgroup$
– egreg
Dec 16 '18 at 14:46
$begingroup$
@egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
$endgroup$
– Galc127
Dec 16 '18 at 14:48
add a comment |
2
$begingroup$
$a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
$endgroup$
– egreg
Dec 16 '18 at 14:46
$begingroup$
@egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
$endgroup$
– Galc127
Dec 16 '18 at 14:48
2
2
$begingroup$
$a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
$endgroup$
– egreg
Dec 16 '18 at 14:46
$begingroup$
$a^2b^2+a^2+ab+b^2=(ab)^2+(a+b/2)^2+3b^2/4$. A sum of squares is nonnegative and is zero if and only if each term is zero.
$endgroup$
– egreg
Dec 16 '18 at 14:46
$begingroup$
@egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
$endgroup$
– Galc127
Dec 16 '18 at 14:48
$begingroup$
@egreg, thanks, I have looked after the factorization, but didn't manage to compute it by myself.
$endgroup$
– Galc127
Dec 16 '18 at 14:48
add a comment |
$begingroup$
Hint: Prove that, for each real $x$, $f'(x)geqslant0$.
$endgroup$
$begingroup$
I did it, but still dont know how it's can help me
$endgroup$
– shay
Dec 16 '18 at 14:08
1
$begingroup$
SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
$endgroup$
– José Carlos Santos
Dec 16 '18 at 14:10
add a comment |
$begingroup$
Hint: Prove that, for each real $x$, $f'(x)geqslant0$.
$endgroup$
$begingroup$
I did it, but still dont know how it's can help me
$endgroup$
– shay
Dec 16 '18 at 14:08
1
$begingroup$
SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
$endgroup$
– José Carlos Santos
Dec 16 '18 at 14:10
add a comment |
$begingroup$
Hint: Prove that, for each real $x$, $f'(x)geqslant0$.
$endgroup$
Hint: Prove that, for each real $x$, $f'(x)geqslant0$.
answered Dec 16 '18 at 14:05
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
$begingroup$
I did it, but still dont know how it's can help me
$endgroup$
– shay
Dec 16 '18 at 14:08
1
$begingroup$
SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
$endgroup$
– José Carlos Santos
Dec 16 '18 at 14:10
add a comment |
$begingroup$
I did it, but still dont know how it's can help me
$endgroup$
– shay
Dec 16 '18 at 14:08
1
$begingroup$
SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
$endgroup$
– José Carlos Santos
Dec 16 '18 at 14:10
$begingroup$
I did it, but still dont know how it's can help me
$endgroup$
– shay
Dec 16 '18 at 14:08
$begingroup$
I did it, but still dont know how it's can help me
$endgroup$
– shay
Dec 16 '18 at 14:08
1
1
$begingroup$
SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
$endgroup$
– José Carlos Santos
Dec 16 '18 at 14:10
$begingroup$
SInce $f'(x)$ is always non-negative, $f$ is invreasing. Now, suppose that $f(x)=f(y)$ for some $x$ and some $y$ such that $x<y$. Then $f$ is constant in $[x,y]$ and therefore $f'$ is $0$ there. But $f'$ only has one single zero.
$endgroup$
– José Carlos Santos
Dec 16 '18 at 14:10
add a comment |
$begingroup$
Here is an approach if you want to avoid derivatives. For $x neq 0$, we have
$$frac{x^3}{x^2 + 1} = frac{1}{frac{1}{x} + frac{1}{x^3}}.$$
The function $1/x + 1/x^3$ for $x > 0$ is the sum of two strictly decreasing functions, hence itself strictly decreasing and therefore one to one with range $(0, infty)$. Hence $1/(1/x + 1/x^3)$ is strictly increasing and one to one with range $(0, infty)$. For $x < 0$, notice $1/x + 1/x^3$ is an odd function, and therefore a similar argument shows $1/(1/x + 1/x^3)$ is one to one with range $(-infty, 0)$.
In fact, this shows that if you add back the point $x = 0$, the function is a bijection since the function is zero here.
$endgroup$
add a comment |
$begingroup$
Here is an approach if you want to avoid derivatives. For $x neq 0$, we have
$$frac{x^3}{x^2 + 1} = frac{1}{frac{1}{x} + frac{1}{x^3}}.$$
The function $1/x + 1/x^3$ for $x > 0$ is the sum of two strictly decreasing functions, hence itself strictly decreasing and therefore one to one with range $(0, infty)$. Hence $1/(1/x + 1/x^3)$ is strictly increasing and one to one with range $(0, infty)$. For $x < 0$, notice $1/x + 1/x^3$ is an odd function, and therefore a similar argument shows $1/(1/x + 1/x^3)$ is one to one with range $(-infty, 0)$.
In fact, this shows that if you add back the point $x = 0$, the function is a bijection since the function is zero here.
$endgroup$
add a comment |
$begingroup$
Here is an approach if you want to avoid derivatives. For $x neq 0$, we have
$$frac{x^3}{x^2 + 1} = frac{1}{frac{1}{x} + frac{1}{x^3}}.$$
The function $1/x + 1/x^3$ for $x > 0$ is the sum of two strictly decreasing functions, hence itself strictly decreasing and therefore one to one with range $(0, infty)$. Hence $1/(1/x + 1/x^3)$ is strictly increasing and one to one with range $(0, infty)$. For $x < 0$, notice $1/x + 1/x^3$ is an odd function, and therefore a similar argument shows $1/(1/x + 1/x^3)$ is one to one with range $(-infty, 0)$.
In fact, this shows that if you add back the point $x = 0$, the function is a bijection since the function is zero here.
$endgroup$
Here is an approach if you want to avoid derivatives. For $x neq 0$, we have
$$frac{x^3}{x^2 + 1} = frac{1}{frac{1}{x} + frac{1}{x^3}}.$$
The function $1/x + 1/x^3$ for $x > 0$ is the sum of two strictly decreasing functions, hence itself strictly decreasing and therefore one to one with range $(0, infty)$. Hence $1/(1/x + 1/x^3)$ is strictly increasing and one to one with range $(0, infty)$. For $x < 0$, notice $1/x + 1/x^3$ is an odd function, and therefore a similar argument shows $1/(1/x + 1/x^3)$ is one to one with range $(-infty, 0)$.
In fact, this shows that if you add back the point $x = 0$, the function is a bijection since the function is zero here.
answered Dec 16 '18 at 14:17
RileyRiley
1825
1825
add a comment |
add a comment |
$begingroup$
Suppose $f(a)=f(b)$: then
$$
a^3(b^2+1)=b^3(a^2+1)
$$
This implies
$$
a^3b^2-a^2b^3+a^3-b^3=0
$$
and so
$$
a^2b^2(a-b)+(a-b)(a^2+ab+b^2)=0
$$
Conclude.
If you know that $a^2+ab+b^2le0$ if and only if $a=b=0$, you're done.
With calculus,
$$
f'(x)=frac{3x^2(x^2+1)-2x^4}{(x^2+1)^2}
$$
The derivative only vanishes at $0$, so Rolle's theorem would only apply to an interval of the form $[a,b]$, with $a<0$ and $b>0$. Can $f(a)=f(b)$ in this case?
$endgroup$
add a comment |
$begingroup$
Suppose $f(a)=f(b)$: then
$$
a^3(b^2+1)=b^3(a^2+1)
$$
This implies
$$
a^3b^2-a^2b^3+a^3-b^3=0
$$
and so
$$
a^2b^2(a-b)+(a-b)(a^2+ab+b^2)=0
$$
Conclude.
If you know that $a^2+ab+b^2le0$ if and only if $a=b=0$, you're done.
With calculus,
$$
f'(x)=frac{3x^2(x^2+1)-2x^4}{(x^2+1)^2}
$$
The derivative only vanishes at $0$, so Rolle's theorem would only apply to an interval of the form $[a,b]$, with $a<0$ and $b>0$. Can $f(a)=f(b)$ in this case?
$endgroup$
add a comment |
$begingroup$
Suppose $f(a)=f(b)$: then
$$
a^3(b^2+1)=b^3(a^2+1)
$$
This implies
$$
a^3b^2-a^2b^3+a^3-b^3=0
$$
and so
$$
a^2b^2(a-b)+(a-b)(a^2+ab+b^2)=0
$$
Conclude.
If you know that $a^2+ab+b^2le0$ if and only if $a=b=0$, you're done.
With calculus,
$$
f'(x)=frac{3x^2(x^2+1)-2x^4}{(x^2+1)^2}
$$
The derivative only vanishes at $0$, so Rolle's theorem would only apply to an interval of the form $[a,b]$, with $a<0$ and $b>0$. Can $f(a)=f(b)$ in this case?
$endgroup$
Suppose $f(a)=f(b)$: then
$$
a^3(b^2+1)=b^3(a^2+1)
$$
This implies
$$
a^3b^2-a^2b^3+a^3-b^3=0
$$
and so
$$
a^2b^2(a-b)+(a-b)(a^2+ab+b^2)=0
$$
Conclude.
If you know that $a^2+ab+b^2le0$ if and only if $a=b=0$, you're done.
With calculus,
$$
f'(x)=frac{3x^2(x^2+1)-2x^4}{(x^2+1)^2}
$$
The derivative only vanishes at $0$, so Rolle's theorem would only apply to an interval of the form $[a,b]$, with $a<0$ and $b>0$. Can $f(a)=f(b)$ in this case?
answered Dec 16 '18 at 14:09
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
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