Finding the minimum value of a function without using Calculus
$begingroup$
Find the minimum value the function $f(x) = x^4 + frac{1}{x^2}$ when $x in Bbb R^*$
My attempt:
Finding the minimum value of this function using calculus is a piece of cake. But since this question appeared on my test when Calculus was not taught to me, there must definitely be a way (Probably by purely using Algebra) to find the minimum value of this function without using Calculus which I am unaware of.
I tried making perfect squares but that got me nowhere. Maybe, I wasn't making the perfect, perfect square :)
Any help would be appreciated.
functions maxima-minima
edited Mar 9 at 14:06
Bernard
123k741116
asked Mar 9 at 13:53
TonyTony
907
$endgroup$
add a comment |
$begingroup$
Find the minimum value the function $f(x) = x^4 + frac{1}{x^2}$ when $x in Bbb R^*$
My attempt:
Finding the minimum value of this function using calculus is a piece of cake. But since this question appeared on my test when Calculus was not taught to me, there must definitely be a way (Probably by purely using Algebra) to find the minimum value of this function without using Calculus which I am unaware of.
I tried making perfect squares but that got me nowhere. Maybe, I wasn't making the perfect, perfect square :)
Any help would be appreciated.
functions maxima-minima
edited Mar 9 at 14:06
Bernard
123k741116
asked Mar 9 at 13:53
TonyTony
907
$endgroup$
add a comment |
$begingroup$
Find the minimum value the function $f(x) = x^4 + frac{1}{x^2}$ when $x in Bbb R^*$
My attempt:
Finding the minimum value of this function using calculus is a piece of cake. But since this question appeared on my test when Calculus was not taught to me, there must definitely be a way (Probably by purely using Algebra) to find the minimum value of this function without using Calculus which I am unaware of.
I tried making perfect squares but that got me nowhere. Maybe, I wasn't making the perfect, perfect square :)
Any help would be appreciated.
functions maxima-minima
edited Mar 9 at 14:06
Bernard
123k741116
asked Mar 9 at 13:53
TonyTony
907
$endgroup$
Find the minimum value the function $f(x) = x^4 + frac{1}{x^2}$ when $x in Bbb R^*$
My attempt:
Finding the minimum value of this function using calculus is a piece of cake. But since this question appeared on my test when Calculus was not taught to me, there must definitely be a way (Probably by purely using Algebra) to find the minimum value of this function without using Calculus which I am unaware of.
I tried making perfect squares but that got me nowhere. Maybe, I wasn't making the perfect, perfect square :)
Any help would be appreciated.
functions maxima-minima
functions maxima-minima
edited Mar 9 at 14:06
Bernard
123k741116
asked Mar 9 at 13:53
TonyTony
907
edited Mar 9 at 14:06
Bernard
123k741116
asked Mar 9 at 13:53
TonyTony
907
edited Mar 9 at 14:06
Bernard
123k741116
edited Mar 9 at 14:06
Bernard
123k741116
edited Mar 9 at 14:06
Bernard
123k741116
123k741116
asked Mar 9 at 13:53
TonyTony
907
asked Mar 9 at 13:53
TonyTony
907
asked Mar 9 at 13:53
TonyTony
907
907
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Use AM-GM:
$$x^4+frac1{x^2}=x^4+frac1{2x^2}+frac1{2x^2}ge 3sqrt[3]{frac1{4}},$$
equality occurs when $x^4=frac1{2x^2}=frac1{2x^2} Rightarrow x=pmfrac1{sqrt[6]{2}}$.
answered Mar 9 at 14:00
farruhotafarruhota
21.1k2841
$endgroup$
add a comment |
$begingroup$
Hint:
As $x^4,1/x^2>0$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality
$$dfrac{ax^4+bx^{-2}}{a+b}gesqrt[a+b]{x^{4a-2b}}$$
Set $4a-2b=0iff b=2a$
answered Mar 9 at 13:58
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
$begingroup$
@farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
$endgroup$
– lab bhattacharjee
Mar 9 at 17:48
add a comment |
$begingroup$
First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + frac{1}{x^2} - b = 0$ has a (double) root. Observe
$$ x^4 + frac{1}{x^2} - b = frac{x^6 - b x^2 + 1}{x^2} $$
has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.
The discriminant of $(x^2)^3 - b(x^2) + 1$ is
$$ -4(-b)^3 - 27 cdot (1)^2 = 4b^3 - 27 text{.} $$
The discriminant is zero if and only if the polynomial has a double root. Taking $b = sqrt[3]{27/4} = frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.
answered Mar 10 at 0:22
Eric TowersEric Towers
32.9k22370
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use AM-GM:
$$x^4+frac1{x^2}=x^4+frac1{2x^2}+frac1{2x^2}ge 3sqrt[3]{frac1{4}},$$
equality occurs when $x^4=frac1{2x^2}=frac1{2x^2} Rightarrow x=pmfrac1{sqrt[6]{2}}$.
answered Mar 9 at 14:00
farruhotafarruhota
21.1k2841
$endgroup$
add a comment |
$begingroup$
Use AM-GM:
$$x^4+frac1{x^2}=x^4+frac1{2x^2}+frac1{2x^2}ge 3sqrt[3]{frac1{4}},$$
equality occurs when $x^4=frac1{2x^2}=frac1{2x^2} Rightarrow x=pmfrac1{sqrt[6]{2}}$.
answered Mar 9 at 14:00
farruhotafarruhota
21.1k2841
$endgroup$
add a comment |
$begingroup$
Use AM-GM:
$$x^4+frac1{x^2}=x^4+frac1{2x^2}+frac1{2x^2}ge 3sqrt[3]{frac1{4}},$$
equality occurs when $x^4=frac1{2x^2}=frac1{2x^2} Rightarrow x=pmfrac1{sqrt[6]{2}}$.
answered Mar 9 at 14:00
farruhotafarruhota
21.1k2841
$endgroup$
Use AM-GM:
$$x^4+frac1{x^2}=x^4+frac1{2x^2}+frac1{2x^2}ge 3sqrt[3]{frac1{4}},$$
equality occurs when $x^4=frac1{2x^2}=frac1{2x^2} Rightarrow x=pmfrac1{sqrt[6]{2}}$.
answered Mar 9 at 14:00
farruhotafarruhota
21.1k2841
answered Mar 9 at 14:00
farruhotafarruhota
21.1k2841
answered Mar 9 at 14:00
farruhotafarruhota
21.1k2841
answered Mar 9 at 14:00
farruhotafarruhota
21.1k2841
21.1k2841
add a comment |
add a comment |
$begingroup$
Hint:
As $x^4,1/x^2>0$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality
$$dfrac{ax^4+bx^{-2}}{a+b}gesqrt[a+b]{x^{4a-2b}}$$
Set $4a-2b=0iff b=2a$
answered Mar 9 at 13:58
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
$begingroup$
@farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
$endgroup$
– lab bhattacharjee
Mar 9 at 17:48
add a comment |
$begingroup$
Hint:
As $x^4,1/x^2>0$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality
$$dfrac{ax^4+bx^{-2}}{a+b}gesqrt[a+b]{x^{4a-2b}}$$
Set $4a-2b=0iff b=2a$
answered Mar 9 at 13:58
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
$begingroup$
@farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
$endgroup$
– lab bhattacharjee
Mar 9 at 17:48
add a comment |
$begingroup$
Hint:
As $x^4,1/x^2>0$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality
$$dfrac{ax^4+bx^{-2}}{a+b}gesqrt[a+b]{x^{4a-2b}}$$
Set $4a-2b=0iff b=2a$
answered Mar 9 at 13:58
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
Hint:
As $x^4,1/x^2>0$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality
$$dfrac{ax^4+bx^{-2}}{a+b}gesqrt[a+b]{x^{4a-2b}}$$
Set $4a-2b=0iff b=2a$
answered Mar 9 at 13:58
lab bhattacharjeelab bhattacharjee
227k15158275
edited Mar 9 at 15:56
answered Mar 9 at 13:58
lab bhattacharjeelab bhattacharjee
227k15158275
answered Mar 9 at 13:58
lab bhattacharjeelab bhattacharjee
227k15158275
answered Mar 9 at 13:58
lab bhattacharjeelab bhattacharjee
227k15158275
227k15158275
$begingroup$
@farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
$endgroup$
– lab bhattacharjee
Mar 9 at 17:48
add a comment |
$begingroup$
@farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
$endgroup$
– lab bhattacharjee
Mar 9 at 17:48
$begingroup$
@farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
$endgroup$
– lab bhattacharjee
Mar 9 at 17:48
$begingroup$
@farruhota, Sorry for the typo. I was too much engrossed with $b=2a$
$endgroup$
– lab bhattacharjee
Mar 9 at 17:48
add a comment |
$begingroup$
First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + frac{1}{x^2} - b = 0$ has a (double) root. Observe
$$ x^4 + frac{1}{x^2} - b = frac{x^6 - b x^2 + 1}{x^2} $$
has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.
The discriminant of $(x^2)^3 - b(x^2) + 1$ is
$$ -4(-b)^3 - 27 cdot (1)^2 = 4b^3 - 27 text{.} $$
The discriminant is zero if and only if the polynomial has a double root. Taking $b = sqrt[3]{27/4} = frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.
answered Mar 10 at 0:22
Eric TowersEric Towers
32.9k22370
$endgroup$
add a comment |
$begingroup$
First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + frac{1}{x^2} - b = 0$ has a (double) root. Observe
$$ x^4 + frac{1}{x^2} - b = frac{x^6 - b x^2 + 1}{x^2} $$
has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.
The discriminant of $(x^2)^3 - b(x^2) + 1$ is
$$ -4(-b)^3 - 27 cdot (1)^2 = 4b^3 - 27 text{.} $$
The discriminant is zero if and only if the polynomial has a double root. Taking $b = sqrt[3]{27/4} = frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.
answered Mar 10 at 0:22
Eric TowersEric Towers
32.9k22370
$endgroup$
add a comment |
$begingroup$
First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + frac{1}{x^2} - b = 0$ has a (double) root. Observe
$$ x^4 + frac{1}{x^2} - b = frac{x^6 - b x^2 + 1}{x^2} $$
has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.
The discriminant of $(x^2)^3 - b(x^2) + 1$ is
$$ -4(-b)^3 - 27 cdot (1)^2 = 4b^3 - 27 text{.} $$
The discriminant is zero if and only if the polynomial has a double root. Taking $b = sqrt[3]{27/4} = frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.
answered Mar 10 at 0:22
Eric TowersEric Towers
32.9k22370
$endgroup$
First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + frac{1}{x^2} - b = 0$ has a (double) root. Observe
$$ x^4 + frac{1}{x^2} - b = frac{x^6 - b x^2 + 1}{x^2} $$
has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.
The discriminant of $(x^2)^3 - b(x^2) + 1$ is
$$ -4(-b)^3 - 27 cdot (1)^2 = 4b^3 - 27 text{.} $$
The discriminant is zero if and only if the polynomial has a double root. Taking $b = sqrt[3]{27/4} = frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.
answered Mar 10 at 0:22
Eric TowersEric Towers
32.9k22370
answered Mar 10 at 0:22
Eric TowersEric Towers
32.9k22370
answered Mar 10 at 0:22
Eric TowersEric Towers
32.9k22370
answered Mar 10 at 0:22
Eric TowersEric Towers
32.9k22370
32.9k22370
add a comment |
add a comment |
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