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Lagrange's Theorem ($f(b)-f(a)=f'(c)(b-a)$) problem

Use Lagrange Remainder Theorem to Prove InequalityProve that $frac{1}{(n+5)^3}$ is a Cauchy sequence.A basic question in the definition of limit pointIncreasing function with dense image continuous?The Zeros Localization Theorem and the Extreme value Theoreminflection points of the solution of a Cauchy problemabout the existence-uniqueness theoremConditions to apply variation of constant méthod on a Cauchy problemIs this seemingly new proof that $[0, 1]$ is compact correct?ODE theorem about solutions related to improper integral

2

2

$begingroup$

Let's consider function $f(x) = sqrt[4]{x}$ on the interval $[a,b]$ with $a,b > 0$. Find point $c in (a,b)$, for which the condition of the Lagrange Theorem holds. (Don't forget to check that the found point is inside of $[a,b]$).

So the Lagrange condition is the following: $f(b)-f(a)=f'(c)(b-a)$.

I have found the solution like this:

$$sqrt[4]{b}-sqrt[4]{a} = frac{1}{4 c^{3/4}}(b-a)$$

Which leads me to the following solution:

$$frac{(b-a)^{4/3}}{(1/4)^{4/3}(b^{1/4}-a^{1/4})^{4/3}}$$

And this apparently is not a correct answer. Plus I have not checked that this point is in $(a,b)$.

Here is how you check:

$$a = left(frac{(sqrt[4]{a} + sqrt[4]{a})(sqrt{a} + sqrt{a})}{4}right)^{4/3} < left(frac{(sqrt[4]{a} + sqrt[4]{b})(sqrt{a} + sqrt{b})}{4}right)^{4/3} < left(frac{(sqrt[4]{b} + sqrt[4]{b})(sqrt{b} + sqrt{b})}{4}right)^{4/3} = b$$

real-analysis

share|cite|improve this question

edited Mar 10 at 14:35

i squared - Keep it Real

asked Mar 10 at 13:58

i squared - Keep it Reali squared - Keep it Real

1,61311027

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Should be $4$ in denominator, not $1/4$.
  $endgroup$
  – Nicolas FRANCOIS
  Mar 10 at 14:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $cin(a,b)$, $f$ is concave, $f'$ is decreasing, and $f'(a)>frac{f(b)-f(a)}{b-a}>f'(b)$, so by IVT applied to $f'$, $c$ must be between $a$ and $b$. Doing it by algebra looks tricky...
  $endgroup$
  – Nicolas FRANCOIS
  Mar 10 at 14:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks @NicolasFRANCOIS
  $endgroup$
  – i squared - Keep it Real
  Mar 10 at 14:31
  

  
  
  
  

add a comment | 

2

2

$begingroup$

Let's consider function $f(x) = sqrt[4]{x}$ on the interval $[a,b]$ with $a,b > 0$. Find point $c in (a,b)$, for which the condition of the Lagrange Theorem holds. (Don't forget to check that the found point is inside of $[a,b]$).

So the Lagrange condition is the following: $f(b)-f(a)=f'(c)(b-a)$.

I have found the solution like this:

$$sqrt[4]{b}-sqrt[4]{a} = frac{1}{4 c^{3/4}}(b-a)$$

Which leads me to the following solution:

$$frac{(b-a)^{4/3}}{(1/4)^{4/3}(b^{1/4}-a^{1/4})^{4/3}}$$

And this apparently is not a correct answer. Plus I have not checked that this point is in $(a,b)$.

Here is how you check:

$$a = left(frac{(sqrt[4]{a} + sqrt[4]{a})(sqrt{a} + sqrt{a})}{4}right)^{4/3} < left(frac{(sqrt[4]{a} + sqrt[4]{b})(sqrt{a} + sqrt{b})}{4}right)^{4/3} < left(frac{(sqrt[4]{b} + sqrt[4]{b})(sqrt{b} + sqrt{b})}{4}right)^{4/3} = b$$

real-analysis

share|cite|improve this question

edited Mar 10 at 14:35

i squared - Keep it Real

asked Mar 10 at 13:58

i squared - Keep it Reali squared - Keep it Real

1,61311027

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Should be $4$ in denominator, not $1/4$.
  $endgroup$
  – Nicolas FRANCOIS
  Mar 10 at 14:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $cin(a,b)$, $f$ is concave, $f'$ is decreasing, and $f'(a)>frac{f(b)-f(a)}{b-a}>f'(b)$, so by IVT applied to $f'$, $c$ must be between $a$ and $b$. Doing it by algebra looks tricky...
  $endgroup$
  – Nicolas FRANCOIS
  Mar 10 at 14:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks @NicolasFRANCOIS
  $endgroup$
  – i squared - Keep it Real
  Mar 10 at 14:31
  

  
  
  
  

add a comment | 

$begingroup$

Let's consider function $f(x) = sqrt[4]{x}$ on the interval $[a,b]$ with $a,b > 0$. Find point $c in (a,b)$, for which the condition of the Lagrange Theorem holds. (Don't forget to check that the found point is inside of $[a,b]$).

So the Lagrange condition is the following: $f(b)-f(a)=f'(c)(b-a)$.

I have found the solution like this:

$$sqrt[4]{b}-sqrt[4]{a} = frac{1}{4 c^{3/4}}(b-a)$$

Which leads me to the following solution:

$$frac{(b-a)^{4/3}}{(1/4)^{4/3}(b^{1/4}-a^{1/4})^{4/3}}$$

And this apparently is not a correct answer. Plus I have not checked that this point is in $(a,b)$.

Here is how you check:

$$a = left(frac{(sqrt[4]{a} + sqrt[4]{a})(sqrt{a} + sqrt{a})}{4}right)^{4/3} < left(frac{(sqrt[4]{a} + sqrt[4]{b})(sqrt{a} + sqrt{b})}{4}right)^{4/3} < left(frac{(sqrt[4]{b} + sqrt[4]{b})(sqrt{b} + sqrt{b})}{4}right)^{4/3} = b$$

real-analysis

share|cite|improve this question

edited Mar 10 at 14:35

i squared - Keep it Real

asked Mar 10 at 13:58

i squared - Keep it Reali squared - Keep it Real

1,61311027

$endgroup$

share|cite|improve this question

edited Mar 10 at 14:35

i squared - Keep it Real

asked Mar 10 at 13:58

i squared - Keep it Reali squared - Keep it Real

1,61311027

share|cite|improve this question

edited Mar 10 at 14:35

i squared - Keep it Real

asked Mar 10 at 13:58

i squared - Keep it Reali squared - Keep it Real

1,61311027

asked Mar 10 at 13:58

i squared - Keep it Reali squared - Keep it Real

1,61311027

asked Mar 10 at 13:58

i squared - Keep it Reali squared - Keep it Real

1,61311027

1 Answer
1

active

oldest

votes

1

$begingroup$

The derivative is $f'(x)=x^{-3/4}/4$, so the point $c$ satisfies
$$
4frac{b^{1/4}-a^{1/4}}{b-a}=c^{-3/4}
$$
Note that the solution is unique. Raise both sides to the power $-4/3$ and find
$$
c=left(frac{b-a}{4(b^{1/4}-a^{1/4})}right)^{!4/3}
$$
Your computation is indeed correct.

There is no need to prove that $a<c<b$, because Lagrange's theorem tells us this holds.

If you have to go with the torture, set $u=a^{1/4}$, $v=b^{1/4}$ and prove that
$$
left(frac{v^4-u^4}{4(v-u)}right)^{!4/3}<v^4
$$
which becomes
$$
frac{v^4-u^4}{4(v-u)}<v^3
$$
that is,
$$
v^3+uv^2+u^2v+u^3<4v^3
$$
Let $u/v=w$, so $0<w<1$. Then the inequality becomes
$$
w^3+w^2+w-3<0
$$
The polynomial factors as $(w-1)(w^2+w+3)$ and the latter factor has no roots. So the only root is $1$ and the polynomial assumes negative values for $w<1$.

Do similarly for the inequality $c>a$.

share|cite|improve this answer

answered Mar 10 at 15:18

egregegreg

184k1486205

$endgroup$

add a comment | 

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1

$begingroup$

The derivative is $f'(x)=x^{-3/4}/4$, so the point $c$ satisfies
$$
4frac{b^{1/4}-a^{1/4}}{b-a}=c^{-3/4}
$$
Note that the solution is unique. Raise both sides to the power $-4/3$ and find
$$
c=left(frac{b-a}{4(b^{1/4}-a^{1/4})}right)^{!4/3}
$$
Your computation is indeed correct.

There is no need to prove that $a<c<b$, because Lagrange's theorem tells us this holds.

If you have to go with the torture, set $u=a^{1/4}$, $v=b^{1/4}$ and prove that
$$
left(frac{v^4-u^4}{4(v-u)}right)^{!4/3}<v^4
$$
which becomes
$$
frac{v^4-u^4}{4(v-u)}<v^3
$$
that is,
$$
v^3+uv^2+u^2v+u^3<4v^3
$$
Let $u/v=w$, so $0<w<1$. Then the inequality becomes
$$
w^3+w^2+w-3<0
$$
The polynomial factors as $(w-1)(w^2+w+3)$ and the latter factor has no roots. So the only root is $1$ and the polynomial assumes negative values for $w<1$.

Do similarly for the inequality $c>a$.

share|cite|improve this answer

answered Mar 10 at 15:18

egregegreg

184k1486205

$endgroup$

add a comment | 

1

$begingroup$

The derivative is $f'(x)=x^{-3/4}/4$, so the point $c$ satisfies
$$
4frac{b^{1/4}-a^{1/4}}{b-a}=c^{-3/4}
$$
Note that the solution is unique. Raise both sides to the power $-4/3$ and find
$$
c=left(frac{b-a}{4(b^{1/4}-a^{1/4})}right)^{!4/3}
$$
Your computation is indeed correct.

There is no need to prove that $a<c<b$, because Lagrange's theorem tells us this holds.

If you have to go with the torture, set $u=a^{1/4}$, $v=b^{1/4}$ and prove that
$$
left(frac{v^4-u^4}{4(v-u)}right)^{!4/3}<v^4
$$
which becomes
$$
frac{v^4-u^4}{4(v-u)}<v^3
$$
that is,
$$
v^3+uv^2+u^2v+u^3<4v^3
$$
Let $u/v=w$, so $0<w<1$. Then the inequality becomes
$$
w^3+w^2+w-3<0
$$
The polynomial factors as $(w-1)(w^2+w+3)$ and the latter factor has no roots. So the only root is $1$ and the polynomial assumes negative values for $w<1$.

Do similarly for the inequality $c>a$.

share|cite|improve this answer

answered Mar 10 at 15:18

egregegreg

184k1486205

$endgroup$

add a comment | 

$begingroup$

The derivative is $f'(x)=x^{-3/4}/4$, so the point $c$ satisfies
$$
4frac{b^{1/4}-a^{1/4}}{b-a}=c^{-3/4}
$$
Note that the solution is unique. Raise both sides to the power $-4/3$ and find
$$
c=left(frac{b-a}{4(b^{1/4}-a^{1/4})}right)^{!4/3}
$$
Your computation is indeed correct.

There is no need to prove that $a<c<b$, because Lagrange's theorem tells us this holds.

If you have to go with the torture, set $u=a^{1/4}$, $v=b^{1/4}$ and prove that
$$
left(frac{v^4-u^4}{4(v-u)}right)^{!4/3}<v^4
$$
which becomes
$$
frac{v^4-u^4}{4(v-u)}<v^3
$$
that is,
$$
v^3+uv^2+u^2v+u^3<4v^3
$$
Let $u/v=w$, so $0<w<1$. Then the inequality becomes
$$
w^3+w^2+w-3<0
$$
The polynomial factors as $(w-1)(w^2+w+3)$ and the latter factor has no roots. So the only root is $1$ and the polynomial assumes negative values for $w<1$.

Do similarly for the inequality $c>a$.

share|cite|improve this answer

answered Mar 10 at 15:18

egregegreg

184k1486205

$endgroup$

share|cite|improve this answer

answered Mar 10 at 15:18

egregegreg

184k1486205

answered Mar 10 at 15:18

egregegreg

184k1486205

answered Mar 10 at 15:18

egregegreg

184k1486205