Mixing
Find all the intervals in which $ -x^4 + x + 3 ge 0 $
How do I find all the intervals in which $$-x^4 + x + 3 ge 0$$?
First of all, I let $f(x) = -x^4 + x + 3$. Then I used the derivative $f'(x) = -4x^3 + 1$ to study its growth.
I used the expression $a^3-b^3 = (a-b)(a^2+ab+b^2)$ to write $f'(x)$ as the following
$$f'(x) = (1 - xsqrt[3]{4})(1 + x sqrt[3]{4} + (x sqrt[3]{4})^2 )$$
from which I clearly have the solution $x = 1/sqrt[3]{4}$.
For $4^frac{2}{3}x^2 + xsqrt[3]{4} + 1$ we have $Delta = 4^frac{2}{3}-4cdot4^frac{2}{3}cdot1 < 0 $ which means that there are no solutions in $mathbb{R}$.
As $f'(2) < 0$ it means that the function is decreasing on $(1/sqrt[3]{4}, infty)$ and increasing on $(-infty, 1/sqrt[3]{4})$.
But was thinking about an approach to use Rolle's theorem but to no avail as I won't be able to pinpoint solutions of the equation.
calculus functions
edited Nov 20 '18 at 12:56
Mattos
2,73021321
asked Nov 20 '18 at 12:42
Eduard6421
17810
add a comment |
How do I find all the intervals in which $$-x^4 + x + 3 ge 0$$?
First of all, I let $f(x) = -x^4 + x + 3$. Then I used the derivative $f'(x) = -4x^3 + 1$ to study its growth.
I used the expression $a^3-b^3 = (a-b)(a^2+ab+b^2)$ to write $f'(x)$ as the following
$$f'(x) = (1 - xsqrt[3]{4})(1 + x sqrt[3]{4} + (x sqrt[3]{4})^2 )$$
from which I clearly have the solution $x = 1/sqrt[3]{4}$.
For $4^frac{2}{3}x^2 + xsqrt[3]{4} + 1$ we have $Delta = 4^frac{2}{3}-4cdot4^frac{2}{3}cdot1 < 0 $ which means that there are no solutions in $mathbb{R}$.
As $f'(2) < 0$ it means that the function is decreasing on $(1/sqrt[3]{4}, infty)$ and increasing on $(-infty, 1/sqrt[3]{4})$.
But was thinking about an approach to use Rolle's theorem but to no avail as I won't be able to pinpoint solutions of the equation.
calculus functions
edited Nov 20 '18 at 12:56
Mattos
2,73021321
asked Nov 20 '18 at 12:42
Eduard6421
17810
To get $ge$, use ge. Similarly, le for $le$. Also, brackets or cdot are better to use than $*$.
– Mattos
Nov 20 '18 at 12:47
Modified :) Thank you!
– Eduard6421
Nov 20 '18 at 12:48
2
What you wrote is correct, but does not really help you. What you really need are the (real) roots of the polynomial. Since you have a forth order polynomial and there are no obvious solutions you need to solve for them numerically or use en.wikipedia.org/wiki/Quartic_function#Ferrari's_solution which isnt too hard for this problem because you do not have a cubic and quadratic term.
– maxmilgram
Nov 20 '18 at 12:52
add a comment |
How do I find all the intervals in which $$-x^4 + x + 3 ge 0$$?
First of all, I let $f(x) = -x^4 + x + 3$. Then I used the derivative $f'(x) = -4x^3 + 1$ to study its growth.
I used the expression $a^3-b^3 = (a-b)(a^2+ab+b^2)$ to write $f'(x)$ as the following
$$f'(x) = (1 - xsqrt[3]{4})(1 + x sqrt[3]{4} + (x sqrt[3]{4})^2 )$$
from which I clearly have the solution $x = 1/sqrt[3]{4}$.
For $4^frac{2}{3}x^2 + xsqrt[3]{4} + 1$ we have $Delta = 4^frac{2}{3}-4cdot4^frac{2}{3}cdot1 < 0 $ which means that there are no solutions in $mathbb{R}$.
As $f'(2) < 0$ it means that the function is decreasing on $(1/sqrt[3]{4}, infty)$ and increasing on $(-infty, 1/sqrt[3]{4})$.
But was thinking about an approach to use Rolle's theorem but to no avail as I won't be able to pinpoint solutions of the equation.
calculus functions
edited Nov 20 '18 at 12:56
Mattos
2,73021321
asked Nov 20 '18 at 12:42
Eduard6421
17810
How do I find all the intervals in which $$-x^4 + x + 3 ge 0$$?
First of all, I let $f(x) = -x^4 + x + 3$. Then I used the derivative $f'(x) = -4x^3 + 1$ to study its growth.
I used the expression $a^3-b^3 = (a-b)(a^2+ab+b^2)$ to write $f'(x)$ as the following
$$f'(x) = (1 - xsqrt[3]{4})(1 + x sqrt[3]{4} + (x sqrt[3]{4})^2 )$$
from which I clearly have the solution $x = 1/sqrt[3]{4}$.
For $4^frac{2}{3}x^2 + xsqrt[3]{4} + 1$ we have $Delta = 4^frac{2}{3}-4cdot4^frac{2}{3}cdot1 < 0 $ which means that there are no solutions in $mathbb{R}$.
As $f'(2) < 0$ it means that the function is decreasing on $(1/sqrt[3]{4}, infty)$ and increasing on $(-infty, 1/sqrt[3]{4})$.
But was thinking about an approach to use Rolle's theorem but to no avail as I won't be able to pinpoint solutions of the equation.
calculus functions
calculus functions
edited Nov 20 '18 at 12:56
Mattos
2,73021321
asked Nov 20 '18 at 12:42
Eduard6421
17810
edited Nov 20 '18 at 12:56
Mattos
2,73021321
asked Nov 20 '18 at 12:42
Eduard6421
17810
edited Nov 20 '18 at 12:56
Mattos
2,73021321
edited Nov 20 '18 at 12:56
Mattos
2,73021321
edited Nov 20 '18 at 12:56
Mattos
2,73021321
2,73021321
asked Nov 20 '18 at 12:42
Eduard6421
17810
asked Nov 20 '18 at 12:42
Eduard6421
17810
asked Nov 20 '18 at 12:42
Eduard6421
17810
17810
To get $ge$, use ge. Similarly, le for $le$. Also, brackets or cdot are better to use than $*$.
– Mattos
Nov 20 '18 at 12:47
Modified :) Thank you!
– Eduard6421
Nov 20 '18 at 12:48
2
What you wrote is correct, but does not really help you. What you really need are the (real) roots of the polynomial. Since you have a forth order polynomial and there are no obvious solutions you need to solve for them numerically or use en.wikipedia.org/wiki/Quartic_function#Ferrari's_solution which isnt too hard for this problem because you do not have a cubic and quadratic term.
– maxmilgram
Nov 20 '18 at 12:52
add a comment |
To get $ge$, use ge. Similarly, le for $le$. Also, brackets or cdot are better to use than $*$.
– Mattos
Nov 20 '18 at 12:47
Modified :) Thank you!
– Eduard6421
Nov 20 '18 at 12:48
2
What you wrote is correct, but does not really help you. What you really need are the (real) roots of the polynomial. Since you have a forth order polynomial and there are no obvious solutions you need to solve for them numerically or use en.wikipedia.org/wiki/Quartic_function#Ferrari's_solution which isnt too hard for this problem because you do not have a cubic and quadratic term.
– maxmilgram
Nov 20 '18 at 12:52
To get $ge$, use ge. Similarly, le for $le$. Also, brackets or cdot are better to use than $*$.
– Mattos
Nov 20 '18 at 12:47
To get $ge$, use ge. Similarly, le for $le$. Also, brackets or cdot are better to use than $*$.
– Mattos
Nov 20 '18 at 12:47
Modified :) Thank you!
– Eduard6421
Nov 20 '18 at 12:48
Modified :) Thank you!
– Eduard6421
Nov 20 '18 at 12:48
2
2
What you wrote is correct, but does not really help you. What you really need are the (real) roots of the polynomial. Since you have a forth order polynomial and there are no obvious solutions you need to solve for them numerically or use en.wikipedia.org/wiki/Quartic_function#Ferrari's_solution which isnt too hard for this problem because you do not have a cubic and quadratic term.
– maxmilgram
Nov 20 '18 at 12:52
What you wrote is correct, but does not really help you. What you really need are the (real) roots of the polynomial. Since you have a forth order polynomial and there are no obvious solutions you need to solve for them numerically or use en.wikipedia.org/wiki/Quartic_function#Ferrari's_solution which isnt too hard for this problem because you do not have a cubic and quadratic term.
– maxmilgram
Nov 20 '18 at 12:52
add a comment |
1 Answer
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You can find the roots of the derivative in a simpler way:
$$-4x^3+1=0implies x^3=frac14$$ and in the reals the cubic root is unique.
Now if we evaluate the polynomial at $dfrac1{sqrt[3]4}$ we find
$$-dfrac1{4sqrt[3]4}+dfrac1{sqrt[3]4}+3,$$ which is positive.
So there is a single interval where the polynomial is positive, defined by the two roots, which are on either side of $dfrac1{sqrt[3]4}$. By trial an error, we see that this interval contains $-1$ and $1$ but not $-2$ nor $2$.
The roots do have a closed-form expression, but it is a little masochistic.
answered Nov 20 '18 at 13:06
Yves Daoust
124k671221
add a comment |
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1 Answer
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You can find the roots of the derivative in a simpler way:
$$-4x^3+1=0implies x^3=frac14$$ and in the reals the cubic root is unique.
Now if we evaluate the polynomial at $dfrac1{sqrt[3]4}$ we find
$$-dfrac1{4sqrt[3]4}+dfrac1{sqrt[3]4}+3,$$ which is positive.
So there is a single interval where the polynomial is positive, defined by the two roots, which are on either side of $dfrac1{sqrt[3]4}$. By trial an error, we see that this interval contains $-1$ and $1$ but not $-2$ nor $2$.
The roots do have a closed-form expression, but it is a little masochistic.
answered Nov 20 '18 at 13:06
Yves Daoust
124k671221
add a comment |
You can find the roots of the derivative in a simpler way:
$$-4x^3+1=0implies x^3=frac14$$ and in the reals the cubic root is unique.
Now if we evaluate the polynomial at $dfrac1{sqrt[3]4}$ we find
$$-dfrac1{4sqrt[3]4}+dfrac1{sqrt[3]4}+3,$$ which is positive.
So there is a single interval where the polynomial is positive, defined by the two roots, which are on either side of $dfrac1{sqrt[3]4}$. By trial an error, we see that this interval contains $-1$ and $1$ but not $-2$ nor $2$.
The roots do have a closed-form expression, but it is a little masochistic.
answered Nov 20 '18 at 13:06
Yves Daoust
124k671221
add a comment |
You can find the roots of the derivative in a simpler way:
$$-4x^3+1=0implies x^3=frac14$$ and in the reals the cubic root is unique.
Now if we evaluate the polynomial at $dfrac1{sqrt[3]4}$ we find
$$-dfrac1{4sqrt[3]4}+dfrac1{sqrt[3]4}+3,$$ which is positive.
So there is a single interval where the polynomial is positive, defined by the two roots, which are on either side of $dfrac1{sqrt[3]4}$. By trial an error, we see that this interval contains $-1$ and $1$ but not $-2$ nor $2$.
The roots do have a closed-form expression, but it is a little masochistic.
answered Nov 20 '18 at 13:06
Yves Daoust
124k671221
You can find the roots of the derivative in a simpler way:
$$-4x^3+1=0implies x^3=frac14$$ and in the reals the cubic root is unique.
Now if we evaluate the polynomial at $dfrac1{sqrt[3]4}$ we find
$$-dfrac1{4sqrt[3]4}+dfrac1{sqrt[3]4}+3,$$ which is positive.
So there is a single interval where the polynomial is positive, defined by the two roots, which are on either side of $dfrac1{sqrt[3]4}$. By trial an error, we see that this interval contains $-1$ and $1$ but not $-2$ nor $2$.
The roots do have a closed-form expression, but it is a little masochistic.
answered Nov 20 '18 at 13:06
Yves Daoust
124k671221
answered Nov 20 '18 at 13:06
Yves Daoust
124k671221
answered Nov 20 '18 at 13:06
Yves Daoust
124k671221
answered Nov 20 '18 at 13:06
Yves Daoust
124k671221
124k671221
add a comment |
add a comment |
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To get $ge$, use ge. Similarly, le for $le$. Also, brackets or cdot are better to use than $*$.
– Mattos
Nov 20 '18 at 12:47
Modified :) Thank you!
– Eduard6421
Nov 20 '18 at 12:48
2
What you wrote is correct, but does not really help you. What you really need are the (real) roots of the polynomial. Since you have a forth order polynomial and there are no obvious solutions you need to solve for them numerically or use en.wikipedia.org/wiki/Quartic_function#Ferrari's_solution which isnt too hard for this problem because you do not have a cubic and quadratic term.
– maxmilgram
Nov 20 '18 at 12:52