Mixing
Prove that the equation has at most 2 solutions [duplicate]
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This question already has an answer here:
Show $x^n+ax+b=0$ has most two solutions
1 answer
For any numbers a, b and and an even natural number n that the equation
$$x^n+ax+b=0$$ has at most 2 solutions.
My attempt: $f'(x)=nx^{n-1}+a$, $n text{ is even} implies n-1 text{ is odd} implies f'(x) text{ is strictly increasing}$.
Now I want to show that $f'(x)$ has exactly 1 root, so that I can apply Rolle's theorem to get the desired result. I think I need to use IVT, but which endpoints of the interval (c,d) should I take?
P.S. Is this still true if n is odd?
real-analysis
asked Jan 26 at 15:08
dxdydzdxdydz
41410
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marked as duplicate by Dietrich Burde, rtybase, metamorphy, saz, Martin R Jan 26 at 22:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Show $x^n+ax+b=0$ has most two solutions
1 answer
For any numbers a, b and and an even natural number n that the equation
$$x^n+ax+b=0$$ has at most 2 solutions.
My attempt: $f'(x)=nx^{n-1}+a$, $n text{ is even} implies n-1 text{ is odd} implies f'(x) text{ is strictly increasing}$.
Now I want to show that $f'(x)$ has exactly 1 root, so that I can apply Rolle's theorem to get the desired result. I think I need to use IVT, but which endpoints of the interval (c,d) should I take?
P.S. Is this still true if n is odd?
real-analysis
asked Jan 26 at 15:08
dxdydzdxdydz
41410
$endgroup$
marked as duplicate by Dietrich Burde, rtybase, metamorphy, saz, Martin R Jan 26 at 22:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
You mean at most two real solutions. The same question is here : math.stackexchange.com/questions/1534686/…
$endgroup$
– ersh
Jan 26 at 15:19
add a comment |
$begingroup$
This question already has an answer here:
Show $x^n+ax+b=0$ has most two solutions
1 answer
For any numbers a, b and and an even natural number n that the equation
$$x^n+ax+b=0$$ has at most 2 solutions.
My attempt: $f'(x)=nx^{n-1}+a$, $n text{ is even} implies n-1 text{ is odd} implies f'(x) text{ is strictly increasing}$.
Now I want to show that $f'(x)$ has exactly 1 root, so that I can apply Rolle's theorem to get the desired result. I think I need to use IVT, but which endpoints of the interval (c,d) should I take?
P.S. Is this still true if n is odd?
real-analysis
asked Jan 26 at 15:08
dxdydzdxdydz
41410
$endgroup$
This question already has an answer here:
Show $x^n+ax+b=0$ has most two solutions
1 answer
For any numbers a, b and and an even natural number n that the equation
$$x^n+ax+b=0$$ has at most 2 solutions.
My attempt: $f'(x)=nx^{n-1}+a$, $n text{ is even} implies n-1 text{ is odd} implies f'(x) text{ is strictly increasing}$.
Now I want to show that $f'(x)$ has exactly 1 root, so that I can apply Rolle's theorem to get the desired result. I think I need to use IVT, but which endpoints of the interval (c,d) should I take?
P.S. Is this still true if n is odd?
This question already has an answer here:
Show $x^n+ax+b=0$ has most two solutions
1 answer
real-analysis
real-analysis
asked Jan 26 at 15:08
dxdydzdxdydz
41410
asked Jan 26 at 15:08
dxdydzdxdydz
41410
asked Jan 26 at 15:08
dxdydzdxdydz
41410
asked Jan 26 at 15:08
dxdydzdxdydz
41410
asked Jan 26 at 15:08
dxdydzdxdydz
41410
41410
marked as duplicate by Dietrich Burde, rtybase, metamorphy, saz, Martin R Jan 26 at 22:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dietrich Burde, rtybase, metamorphy, saz, Martin R Jan 26 at 22:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
You mean at most two real solutions. The same question is here : math.stackexchange.com/questions/1534686/…
$endgroup$
– ersh
Jan 26 at 15:19
add a comment |
$begingroup$
You mean at most two real solutions. The same question is here : math.stackexchange.com/questions/1534686/…
$endgroup$
– ersh
Jan 26 at 15:19
$begingroup$
You mean at most two real solutions. The same question is here : math.stackexchange.com/questions/1534686/…
$endgroup$
– ersh
Jan 26 at 15:19
$begingroup$
You mean at most two real solutions. The same question is here : math.stackexchange.com/questions/1534686/…
$endgroup$
– ersh
Jan 26 at 15:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$x^3-x$ has three roots.
To your other question, you don't need to show that $f'(x)$ has exactly one root, just that it can't have more than one (which is clear, since it is strictly increasing). It is true, though, that it always has exactly one root...to see that note that $f'(x)$ tends to $-infty$ if $x$ tends to $-infty$ and $f'(x)$ tends to $+infty$ if $x$ tends to $+infty$. More broadly, every odd degree polynomial has at least one real root.
answered Jan 26 at 15:16
lulululu
43.1k25080
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N.B. the Q. states even exponent for the leading term.
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– stochasticboy321
Jan 26 at 15:19
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@stochasticboy321 In the last line, the OP asks if it can be extended to odd $n$.
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– lulu
Jan 26 at 15:38
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@lulu $f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
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– dxdydz
Jan 26 at 16:08
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That is correct.
$endgroup$
– lulu
Jan 26 at 18:03
add a comment |
$begingroup$
Let $$f(x)=x^n+ax+b$$ then $$f(0)=b$$ so we get $$f'(x)=nx^{n-1}+a$$ so we have $$f'(x)geq 0$$ if $$ageq 0$$.
If $$a<0$$ so we obtain $$f'(x)=0$$ if $$x=left(-frac{a}{n}right)^{1/(n-1)}$$
Can you finish?
answered Jan 26 at 15:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
$begingroup$
$f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:04
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$x^3-x$ has three roots.
To your other question, you don't need to show that $f'(x)$ has exactly one root, just that it can't have more than one (which is clear, since it is strictly increasing). It is true, though, that it always has exactly one root...to see that note that $f'(x)$ tends to $-infty$ if $x$ tends to $-infty$ and $f'(x)$ tends to $+infty$ if $x$ tends to $+infty$. More broadly, every odd degree polynomial has at least one real root.
answered Jan 26 at 15:16
lulululu
43.1k25080
$endgroup$
$begingroup$
N.B. the Q. states even exponent for the leading term.
$endgroup$
– stochasticboy321
Jan 26 at 15:19
$begingroup$
@stochasticboy321 In the last line, the OP asks if it can be extended to odd $n$.
$endgroup$
– lulu
Jan 26 at 15:38
$begingroup$
@lulu $f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:08
$begingroup$
That is correct.
$endgroup$
– lulu
Jan 26 at 18:03
add a comment |
$begingroup$
$x^3-x$ has three roots.
To your other question, you don't need to show that $f'(x)$ has exactly one root, just that it can't have more than one (which is clear, since it is strictly increasing). It is true, though, that it always has exactly one root...to see that note that $f'(x)$ tends to $-infty$ if $x$ tends to $-infty$ and $f'(x)$ tends to $+infty$ if $x$ tends to $+infty$. More broadly, every odd degree polynomial has at least one real root.
answered Jan 26 at 15:16
lulululu
43.1k25080
$endgroup$
$begingroup$
N.B. the Q. states even exponent for the leading term.
$endgroup$
– stochasticboy321
Jan 26 at 15:19
$begingroup$
@stochasticboy321 In the last line, the OP asks if it can be extended to odd $n$.
$endgroup$
– lulu
Jan 26 at 15:38
$begingroup$
@lulu $f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:08
$begingroup$
That is correct.
$endgroup$
– lulu
Jan 26 at 18:03
add a comment |
$begingroup$
$x^3-x$ has three roots.
To your other question, you don't need to show that $f'(x)$ has exactly one root, just that it can't have more than one (which is clear, since it is strictly increasing). It is true, though, that it always has exactly one root...to see that note that $f'(x)$ tends to $-infty$ if $x$ tends to $-infty$ and $f'(x)$ tends to $+infty$ if $x$ tends to $+infty$. More broadly, every odd degree polynomial has at least one real root.
answered Jan 26 at 15:16
lulululu
43.1k25080
$endgroup$
$x^3-x$ has three roots.
To your other question, you don't need to show that $f'(x)$ has exactly one root, just that it can't have more than one (which is clear, since it is strictly increasing). It is true, though, that it always has exactly one root...to see that note that $f'(x)$ tends to $-infty$ if $x$ tends to $-infty$ and $f'(x)$ tends to $+infty$ if $x$ tends to $+infty$. More broadly, every odd degree polynomial has at least one real root.
answered Jan 26 at 15:16
lulululu
43.1k25080
answered Jan 26 at 15:16
lulululu
43.1k25080
answered Jan 26 at 15:16
lulululu
43.1k25080
answered Jan 26 at 15:16
lulululu
43.1k25080
43.1k25080
$begingroup$
N.B. the Q. states even exponent for the leading term.
$endgroup$
– stochasticboy321
Jan 26 at 15:19
$begingroup$
@stochasticboy321 In the last line, the OP asks if it can be extended to odd $n$.
$endgroup$
– lulu
Jan 26 at 15:38
$begingroup$
@lulu $f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:08
$begingroup$
That is correct.
$endgroup$
– lulu
Jan 26 at 18:03
add a comment |
$begingroup$
N.B. the Q. states even exponent for the leading term.
$endgroup$
– stochasticboy321
Jan 26 at 15:19
$begingroup$
@stochasticboy321 In the last line, the OP asks if it can be extended to odd $n$.
$endgroup$
– lulu
Jan 26 at 15:38
$begingroup$
@lulu $f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:08
$begingroup$
That is correct.
$endgroup$
– lulu
Jan 26 at 18:03
$begingroup$
N.B. the Q. states even exponent for the leading term.
$endgroup$
– stochasticboy321
Jan 26 at 15:19
$begingroup$
N.B. the Q. states even exponent for the leading term.
$endgroup$
– stochasticboy321
Jan 26 at 15:19
$begingroup$
@stochasticboy321 In the last line, the OP asks if it can be extended to odd $n$.
$endgroup$
– lulu
Jan 26 at 15:38
$begingroup$
@stochasticboy321 In the last line, the OP asks if it can be extended to odd $n$.
$endgroup$
– lulu
Jan 26 at 15:38
$begingroup$
@lulu $f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:08
$begingroup$
@lulu $f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:08
$begingroup$
That is correct.
$endgroup$
– lulu
Jan 26 at 18:03
$begingroup$
That is correct.
$endgroup$
– lulu
Jan 26 at 18:03
add a comment |
$begingroup$
Let $$f(x)=x^n+ax+b$$ then $$f(0)=b$$ so we get $$f'(x)=nx^{n-1}+a$$ so we have $$f'(x)geq 0$$ if $$ageq 0$$.
If $$a<0$$ so we obtain $$f'(x)=0$$ if $$x=left(-frac{a}{n}right)^{1/(n-1)}$$
Can you finish?
answered Jan 26 at 15:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
$begingroup$
$f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:04
add a comment |
$begingroup$
Let $$f(x)=x^n+ax+b$$ then $$f(0)=b$$ so we get $$f'(x)=nx^{n-1}+a$$ so we have $$f'(x)geq 0$$ if $$ageq 0$$.
If $$a<0$$ so we obtain $$f'(x)=0$$ if $$x=left(-frac{a}{n}right)^{1/(n-1)}$$
Can you finish?
answered Jan 26 at 15:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
$begingroup$
$f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:04
add a comment |
$begingroup$
Let $$f(x)=x^n+ax+b$$ then $$f(0)=b$$ so we get $$f'(x)=nx^{n-1}+a$$ so we have $$f'(x)geq 0$$ if $$ageq 0$$.
If $$a<0$$ so we obtain $$f'(x)=0$$ if $$x=left(-frac{a}{n}right)^{1/(n-1)}$$
Can you finish?
answered Jan 26 at 15:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
Let $$f(x)=x^n+ax+b$$ then $$f(0)=b$$ so we get $$f'(x)=nx^{n-1}+a$$ so we have $$f'(x)geq 0$$ if $$ageq 0$$.
If $$a<0$$ so we obtain $$f'(x)=0$$ if $$x=left(-frac{a}{n}right)^{1/(n-1)}$$
Can you finish?
answered Jan 26 at 15:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered Jan 26 at 15:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered Jan 26 at 15:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered Jan 26 at 15:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
78k42866
$begingroup$
$f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:04
add a comment |
$begingroup$
$f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:04
$begingroup$
$f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:04
$begingroup$
$f′(x)$ is strictly increasing, so it has at most 1 root (which is $x=left(-frac{a}{n}right)^{1/(n-1)})$, so by Rolle's theorem $f(x)$ has at most 2 roots, correct?
$endgroup$
– dxdydz
Jan 26 at 16:04
add a comment |
$begingroup$
You mean at most two real solutions. The same question is here : math.stackexchange.com/questions/1534686/…
$endgroup$
– ersh
Jan 26 at 15:19