Proof that $f(x)=4x^4-2x+1$ has no real roots.
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My thought was to:
1) hypothesis there are 2 real roots for this equation,
2) apply Rolle's theorem and come to a reductio ad absurdum
and then if there aren't 2 real roots, it has to be 1. If there is 1 real root, this means that it has to have 3 non-real roots. But non real roots come in pairs, so either is 2 real- 2 non real, either 0 real- 4 non real. Therefore, it has no real roots.
In case my thought is correct, the problem is that $f'(x)=0 => x^3=1/8$ doesn't lead me in reductio ad absurdum, because it has 1 real and 2 non- real roots.
I'm stuck and I'm about to punch the desk. Please, release me from this martyrdom.
calculus
asked Sep 8 at 12:09
Donna Caligine
356
add a comment |Â
up vote
0
down vote
favorite
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My thought was to:
1) hypothesis there are 2 real roots for this equation,
2) apply Rolle's theorem and come to a reductio ad absurdum
and then if there aren't 2 real roots, it has to be 1. If there is 1 real root, this means that it has to have 3 non-real roots. But non real roots come in pairs, so either is 2 real- 2 non real, either 0 real- 4 non real. Therefore, it has no real roots.
In case my thought is correct, the problem is that $f'(x)=0 => x^3=1/8$ doesn't lead me in reductio ad absurdum, because it has 1 real and 2 non- real roots.
I'm stuck and I'm about to punch the desk. Please, release me from this martyrdom.
calculus
asked Sep 8 at 12:09
Donna Caligine
356
And why do you demand a reductio ad absurdum?
– Bernard
Sep 8 at 12:15
@Bernard, my bad actually. It was the only thing that came into my mind at first place and then I was desperately trying to solve it this way.
– Donna Caligine
Sep 8 at 12:29
add a comment |Â
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
My thought was to:
1) hypothesis there are 2 real roots for this equation,
2) apply Rolle's theorem and come to a reductio ad absurdum
and then if there aren't 2 real roots, it has to be 1. If there is 1 real root, this means that it has to have 3 non-real roots. But non real roots come in pairs, so either is 2 real- 2 non real, either 0 real- 4 non real. Therefore, it has no real roots.
In case my thought is correct, the problem is that $f'(x)=0 => x^3=1/8$ doesn't lead me in reductio ad absurdum, because it has 1 real and 2 non- real roots.
I'm stuck and I'm about to punch the desk. Please, release me from this martyrdom.
calculus
asked Sep 8 at 12:09
Donna Caligine
356
My thought was to:
1) hypothesis there are 2 real roots for this equation,
2) apply Rolle's theorem and come to a reductio ad absurdum
and then if there aren't 2 real roots, it has to be 1. If there is 1 real root, this means that it has to have 3 non-real roots. But non real roots come in pairs, so either is 2 real- 2 non real, either 0 real- 4 non real. Therefore, it has no real roots.
In case my thought is correct, the problem is that $f'(x)=0 => x^3=1/8$ doesn't lead me in reductio ad absurdum, because it has 1 real and 2 non- real roots.
I'm stuck and I'm about to punch the desk. Please, release me from this martyrdom.
calculus
calculus
asked Sep 8 at 12:09
Donna Caligine
356
asked Sep 8 at 12:09
Donna Caligine
356
asked Sep 8 at 12:09
Donna Caligine
356
asked Sep 8 at 12:09
Donna Caligine
356
asked Sep 8 at 12:09
Donna Caligine
356
356
And why do you demand a reductio ad absurdum?
– Bernard
Sep 8 at 12:15
@Bernard, my bad actually. It was the only thing that came into my mind at first place and then I was desperately trying to solve it this way.
– Donna Caligine
Sep 8 at 12:29
add a comment |Â
And why do you demand a reductio ad absurdum?
– Bernard
Sep 8 at 12:15
@Bernard, my bad actually. It was the only thing that came into my mind at first place and then I was desperately trying to solve it this way.
– Donna Caligine
Sep 8 at 12:29
And why do you demand a reductio ad absurdum?
– Bernard
Sep 8 at 12:15
And why do you demand a reductio ad absurdum?
– Bernard
Sep 8 at 12:15
@Bernard, my bad actually. It was the only thing that came into my mind at first place and then I was desperately trying to solve it this way.
– Donna Caligine
Sep 8 at 12:29
@Bernard, my bad actually. It was the only thing that came into my mind at first place and then I was desperately trying to solve it this way.
– Donna Caligine
Sep 8 at 12:29
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
- $f'(x)=2(8x^3-1)$, so there's a single critical point: $; x=frac12$.
- $f''(x)=48x^2ge 0$, so by the second derivative test, this critical point is a minimum, and this minimum is an absolute minimum.
- $f(frac12)=frac14>0$.
answered Sep 8 at 12:20
Bernard
112k635104
Actually there is no need to compute the second derivative in this case (although it can't hurt): as function is continuous and diverges to $+infty$ for $x$ going both to $pminfty$ there has to be at least one minimum. As there is only one critical point, it must be that one.
– b00n heT
Sep 8 at 12:26
1
@b00nheT: That's right, but it more or less supposes some familiarity with the general look of quartic function curves. I preferred some general conclusive evidence.
– Bernard
Sep 8 at 12:49
Of course @Bernard. It's just that sometimes I feel like the geometry of the curve gets completely neglected due to these automatic "sign of derivative" approaches :)
– b00n heT
Sep 8 at 15:54
1
I share your point of view., and I must say I first thought exactly like you. Then I wondered at which level was supposed to be asked.
– Bernard
Sep 8 at 16:01
add a comment |Â
up vote
2
down vote
Hint: Use the derivative to find the minimum.
answered Sep 8 at 12:12
b00n heT
8,90211833
Oh thank you so much.
– Donna Caligine
Sep 8 at 12:20
add a comment |Â
up vote
1
down vote
Hint: Compute $$f(1/2)=4(1/2)^4-2(1/2)+1=frac14>0$$
edited Sep 8 at 12:31
amWhy
190k27221433
answered Sep 8 at 12:13
Dr. Sonnhard Graubner
69k32761
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
- $f'(x)=2(8x^3-1)$, so there's a single critical point: $; x=frac12$.
- $f''(x)=48x^2ge 0$, so by the second derivative test, this critical point is a minimum, and this minimum is an absolute minimum.
- $f(frac12)=frac14>0$.
answered Sep 8 at 12:20
Bernard
112k635104
Actually there is no need to compute the second derivative in this case (although it can't hurt): as function is continuous and diverges to $+infty$ for $x$ going both to $pminfty$ there has to be at least one minimum. As there is only one critical point, it must be that one.
– b00n heT
Sep 8 at 12:26
1
@b00nheT: That's right, but it more or less supposes some familiarity with the general look of quartic function curves. I preferred some general conclusive evidence.
– Bernard
Sep 8 at 12:49
Of course @Bernard. It's just that sometimes I feel like the geometry of the curve gets completely neglected due to these automatic "sign of derivative" approaches :)
– b00n heT
Sep 8 at 15:54
1
I share your point of view., and I must say I first thought exactly like you. Then I wondered at which level was supposed to be asked.
– Bernard
Sep 8 at 16:01
add a comment |Â
up vote
6
down vote
accepted
- $f'(x)=2(8x^3-1)$, so there's a single critical point: $; x=frac12$.
- $f''(x)=48x^2ge 0$, so by the second derivative test, this critical point is a minimum, and this minimum is an absolute minimum.
- $f(frac12)=frac14>0$.
answered Sep 8 at 12:20
Bernard
112k635104
Actually there is no need to compute the second derivative in this case (although it can't hurt): as function is continuous and diverges to $+infty$ for $x$ going both to $pminfty$ there has to be at least one minimum. As there is only one critical point, it must be that one.
– b00n heT
Sep 8 at 12:26
1
@b00nheT: That's right, but it more or less supposes some familiarity with the general look of quartic function curves. I preferred some general conclusive evidence.
– Bernard
Sep 8 at 12:49
Of course @Bernard. It's just that sometimes I feel like the geometry of the curve gets completely neglected due to these automatic "sign of derivative" approaches :)
– b00n heT
Sep 8 at 15:54
1
I share your point of view., and I must say I first thought exactly like you. Then I wondered at which level was supposed to be asked.
– Bernard
Sep 8 at 16:01
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
- $f'(x)=2(8x^3-1)$, so there's a single critical point: $; x=frac12$.
- $f''(x)=48x^2ge 0$, so by the second derivative test, this critical point is a minimum, and this minimum is an absolute minimum.
- $f(frac12)=frac14>0$.
answered Sep 8 at 12:20
Bernard
112k635104
- $f'(x)=2(8x^3-1)$, so there's a single critical point: $; x=frac12$.
- $f''(x)=48x^2ge 0$, so by the second derivative test, this critical point is a minimum, and this minimum is an absolute minimum.
- $f(frac12)=frac14>0$.
answered Sep 8 at 12:20
Bernard
112k635104
answered Sep 8 at 12:20
Bernard
112k635104
answered Sep 8 at 12:20
Bernard
112k635104
answered Sep 8 at 12:20
Bernard
112k635104
112k635104
Actually there is no need to compute the second derivative in this case (although it can't hurt): as function is continuous and diverges to $+infty$ for $x$ going both to $pminfty$ there has to be at least one minimum. As there is only one critical point, it must be that one.
– b00n heT
Sep 8 at 12:26
1
@b00nheT: That's right, but it more or less supposes some familiarity with the general look of quartic function curves. I preferred some general conclusive evidence.
– Bernard
Sep 8 at 12:49
Of course @Bernard. It's just that sometimes I feel like the geometry of the curve gets completely neglected due to these automatic "sign of derivative" approaches :)
– b00n heT
Sep 8 at 15:54
1
I share your point of view., and I must say I first thought exactly like you. Then I wondered at which level was supposed to be asked.
– Bernard
Sep 8 at 16:01
add a comment |Â
Actually there is no need to compute the second derivative in this case (although it can't hurt): as function is continuous and diverges to $+infty$ for $x$ going both to $pminfty$ there has to be at least one minimum. As there is only one critical point, it must be that one.
– b00n heT
Sep 8 at 12:26
1
@b00nheT: That's right, but it more or less supposes some familiarity with the general look of quartic function curves. I preferred some general conclusive evidence.
– Bernard
Sep 8 at 12:49
Of course @Bernard. It's just that sometimes I feel like the geometry of the curve gets completely neglected due to these automatic "sign of derivative" approaches :)
– b00n heT
Sep 8 at 15:54
1
I share your point of view., and I must say I first thought exactly like you. Then I wondered at which level was supposed to be asked.
– Bernard
Sep 8 at 16:01
Actually there is no need to compute the second derivative in this case (although it can't hurt): as function is continuous and diverges to $+infty$ for $x$ going both to $pminfty$ there has to be at least one minimum. As there is only one critical point, it must be that one.
– b00n heT
Sep 8 at 12:26
Actually there is no need to compute the second derivative in this case (although it can't hurt): as function is continuous and diverges to $+infty$ for $x$ going both to $pminfty$ there has to be at least one minimum. As there is only one critical point, it must be that one.
– b00n heT
Sep 8 at 12:26
1
1
@b00nheT: That's right, but it more or less supposes some familiarity with the general look of quartic function curves. I preferred some general conclusive evidence.
– Bernard
Sep 8 at 12:49
@b00nheT: That's right, but it more or less supposes some familiarity with the general look of quartic function curves. I preferred some general conclusive evidence.
– Bernard
Sep 8 at 12:49
Of course @Bernard. It's just that sometimes I feel like the geometry of the curve gets completely neglected due to these automatic "sign of derivative" approaches :)
– b00n heT
Sep 8 at 15:54
Of course @Bernard. It's just that sometimes I feel like the geometry of the curve gets completely neglected due to these automatic "sign of derivative" approaches :)
– b00n heT
Sep 8 at 15:54
1
1
I share your point of view., and I must say I first thought exactly like you. Then I wondered at which level was supposed to be asked.
– Bernard
Sep 8 at 16:01
I share your point of view., and I must say I first thought exactly like you. Then I wondered at which level was supposed to be asked.
– Bernard
Sep 8 at 16:01
add a comment |Â
up vote
2
down vote
Hint: Use the derivative to find the minimum.
answered Sep 8 at 12:12
b00n heT
8,90211833
Oh thank you so much.
– Donna Caligine
Sep 8 at 12:20
add a comment |Â
up vote
2
down vote
Hint: Use the derivative to find the minimum.
answered Sep 8 at 12:12
b00n heT
8,90211833
Oh thank you so much.
– Donna Caligine
Sep 8 at 12:20
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: Use the derivative to find the minimum.
answered Sep 8 at 12:12
b00n heT
8,90211833
Hint: Use the derivative to find the minimum.
answered Sep 8 at 12:12
b00n heT
8,90211833
answered Sep 8 at 12:12
b00n heT
8,90211833
answered Sep 8 at 12:12
b00n heT
8,90211833
answered Sep 8 at 12:12
b00n heT
8,90211833
8,90211833
Oh thank you so much.
– Donna Caligine
Sep 8 at 12:20
add a comment |Â
Oh thank you so much.
– Donna Caligine
Sep 8 at 12:20
Oh thank you so much.
– Donna Caligine
Sep 8 at 12:20
Oh thank you so much.
– Donna Caligine
Sep 8 at 12:20
add a comment |Â
up vote
1
down vote
Hint: Compute $$f(1/2)=4(1/2)^4-2(1/2)+1=frac14>0$$
edited Sep 8 at 12:31
amWhy
190k27221433
answered Sep 8 at 12:13
Dr. Sonnhard Graubner
69k32761
add a comment |Â
up vote
1
down vote
Hint: Compute $$f(1/2)=4(1/2)^4-2(1/2)+1=frac14>0$$
edited Sep 8 at 12:31
amWhy
190k27221433
answered Sep 8 at 12:13
Dr. Sonnhard Graubner
69k32761
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Compute $$f(1/2)=4(1/2)^4-2(1/2)+1=frac14>0$$
edited Sep 8 at 12:31
amWhy
190k27221433
answered Sep 8 at 12:13
Dr. Sonnhard Graubner
69k32761
Hint: Compute $$f(1/2)=4(1/2)^4-2(1/2)+1=frac14>0$$
edited Sep 8 at 12:31
amWhy
190k27221433
answered Sep 8 at 12:13
Dr. Sonnhard Graubner
69k32761
edited Sep 8 at 12:31
amWhy
190k27221433
edited Sep 8 at 12:31
amWhy
190k27221433
edited Sep 8 at 12:31
amWhy
190k27221433
190k27221433
answered Sep 8 at 12:13
Dr. Sonnhard Graubner
69k32761
answered Sep 8 at 12:13
Dr. Sonnhard Graubner
69k32761
answered Sep 8 at 12:13
Dr. Sonnhard Graubner
69k32761
69k32761
add a comment |Â
add a comment |Â
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And why do you demand a reductio ad absurdum?
– Bernard
Sep 8 at 12:15
@Bernard, my bad actually. It was the only thing that came into my mind at first place and then I was desperately trying to solve it this way.
– Donna Caligine
Sep 8 at 12:29