Mixing
Real Root of a Polynomial on a closed interval
can you give me some tips to solve this question? 
real-analysis analysis
asked Nov 20 '18 at 21:13
bebe
146
add a comment |
can you give me some tips to solve this question?
real-analysis analysis
asked Nov 20 '18 at 21:13
bebe
146
add a comment |
can you give me some tips to solve this question?
real-analysis analysis
asked Nov 20 '18 at 21:13
bebe
146
can you give me some tips to solve this question?
real-analysis analysis
real-analysis analysis
asked Nov 20 '18 at 21:13
bebe
146
asked Nov 20 '18 at 21:13
bebe
146
asked Nov 20 '18 at 21:13
bebe
146
asked Nov 20 '18 at 21:13
bebe
146
asked Nov 20 '18 at 21:13
bebe
146
146
add a comment |
add a comment |
2 Answers
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I think the simplest way of looking at this would be to construct a second polynomial $Q(x)$ that is, essentially, the antiderivative of $P(x)$ (with a constant term of $C=0$). (Depending on where you are in your coursework, you don't have to say it's the antiderivative to $P$, just say "let $Q(x)$ be ..." It's equally valid.)
Thus, $Q'(x) = P(x)$. From there, apply the mean value theorem to $Q$ on the interval $[0,1]$.
That should be a sufficient nudge forward.
answered Nov 20 '18 at 21:19
Eevee Trainer
4,7201634
add a comment |
Hint: Consider the polynomial
begin{align}
f(x) = C_0 x+frac{C_1}{2}x^2+ldots+frac{C_n}{n+1}x^{n+1}
end{align}
and use Rolle's theorem.
answered Nov 20 '18 at 21:19
Jacky Chong
17.8k21128
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
I think the simplest way of looking at this would be to construct a second polynomial $Q(x)$ that is, essentially, the antiderivative of $P(x)$ (with a constant term of $C=0$). (Depending on where you are in your coursework, you don't have to say it's the antiderivative to $P$, just say "let $Q(x)$ be ..." It's equally valid.)
Thus, $Q'(x) = P(x)$. From there, apply the mean value theorem to $Q$ on the interval $[0,1]$.
That should be a sufficient nudge forward.
answered Nov 20 '18 at 21:19
Eevee Trainer
4,7201634
add a comment |
I think the simplest way of looking at this would be to construct a second polynomial $Q(x)$ that is, essentially, the antiderivative of $P(x)$ (with a constant term of $C=0$). (Depending on where you are in your coursework, you don't have to say it's the antiderivative to $P$, just say "let $Q(x)$ be ..." It's equally valid.)
Thus, $Q'(x) = P(x)$. From there, apply the mean value theorem to $Q$ on the interval $[0,1]$.
That should be a sufficient nudge forward.
answered Nov 20 '18 at 21:19
Eevee Trainer
4,7201634
add a comment |
I think the simplest way of looking at this would be to construct a second polynomial $Q(x)$ that is, essentially, the antiderivative of $P(x)$ (with a constant term of $C=0$). (Depending on where you are in your coursework, you don't have to say it's the antiderivative to $P$, just say "let $Q(x)$ be ..." It's equally valid.)
Thus, $Q'(x) = P(x)$. From there, apply the mean value theorem to $Q$ on the interval $[0,1]$.
That should be a sufficient nudge forward.
answered Nov 20 '18 at 21:19
Eevee Trainer
4,7201634
I think the simplest way of looking at this would be to construct a second polynomial $Q(x)$ that is, essentially, the antiderivative of $P(x)$ (with a constant term of $C=0$). (Depending on where you are in your coursework, you don't have to say it's the antiderivative to $P$, just say "let $Q(x)$ be ..." It's equally valid.)
Thus, $Q'(x) = P(x)$. From there, apply the mean value theorem to $Q$ on the interval $[0,1]$.
That should be a sufficient nudge forward.
answered Nov 20 '18 at 21:19
Eevee Trainer
4,7201634
answered Nov 20 '18 at 21:19
Eevee Trainer
4,7201634
answered Nov 20 '18 at 21:19
Eevee Trainer
4,7201634
answered Nov 20 '18 at 21:19
Eevee Trainer
4,7201634
4,7201634
add a comment |
add a comment |
Hint: Consider the polynomial
begin{align}
f(x) = C_0 x+frac{C_1}{2}x^2+ldots+frac{C_n}{n+1}x^{n+1}
end{align}
and use Rolle's theorem.
answered Nov 20 '18 at 21:19
Jacky Chong
17.8k21128
add a comment |
Hint: Consider the polynomial
begin{align}
f(x) = C_0 x+frac{C_1}{2}x^2+ldots+frac{C_n}{n+1}x^{n+1}
end{align}
and use Rolle's theorem.
answered Nov 20 '18 at 21:19
Jacky Chong
17.8k21128
add a comment |
Hint: Consider the polynomial
begin{align}
f(x) = C_0 x+frac{C_1}{2}x^2+ldots+frac{C_n}{n+1}x^{n+1}
end{align}
and use Rolle's theorem.
answered Nov 20 '18 at 21:19
Jacky Chong
17.8k21128
Hint: Consider the polynomial
begin{align}
f(x) = C_0 x+frac{C_1}{2}x^2+ldots+frac{C_n}{n+1}x^{n+1}
end{align}
and use Rolle's theorem.
answered Nov 20 '18 at 21:19
Jacky Chong
17.8k21128
answered Nov 20 '18 at 21:19
Jacky Chong
17.8k21128
answered Nov 20 '18 at 21:19
Jacky Chong
17.8k21128
answered Nov 20 '18 at 21:19
Jacky Chong
17.8k21128
17.8k21128
add a comment |
add a comment |
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