Maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$What's the minimum value?Finding the maximum and minimum values of $f(x)=a^x+a^1/x$notation for minimum and maximum?Piecewise function and maximum/minMinimum of a maximum functionMaximum value of function given minimum valueMaximum and minimum for piecewise functionMaximum and minimum values of functionStrictly convex minimum and maximumFinding the minimum value of a function without using Calculus
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Maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$
What's the minimum value?Finding the maximum and minimum values of $f(x)=a^x+a^1/x$notation for minimum and maximum?Piecewise function and maximum/minMinimum of a maximum functionMaximum value of function given minimum valueMaximum and minimum for piecewise functionMaximum and minimum values of functionStrictly convex minimum and maximumFinding the minimum value of a function without using Calculus
$begingroup$
Prove that the maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$ are those for which $(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square.
I tried to solve this problem using the value of $y'=0$.
Solving the I got $y=fracax+bAx+B$.
Now I need to prove that $T=(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square by substituting the value of $y=fracax+bAx+B$ but failed.
functions
asked Mar 13 at 10:44
Samar Imam ZaidiSamar Imam Zaidi
1,5751520
$endgroup$
add a comment |
$begingroup$
Prove that the maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$ are those for which $(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square.
I tried to solve this problem using the value of $y'=0$.
Solving the I got $y=fracax+bAx+B$.
Now I need to prove that $T=(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square by substituting the value of $y=fracax+bAx+B$ but failed.
functions
asked Mar 13 at 10:44
Samar Imam ZaidiSamar Imam Zaidi
1,5751520
$endgroup$
$begingroup$
For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
$endgroup$
– Dave L. Renfro
Mar 13 at 11:08
add a comment |
$begingroup$
Prove that the maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$ are those for which $(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square.
I tried to solve this problem using the value of $y'=0$.
Solving the I got $y=fracax+bAx+B$.
Now I need to prove that $T=(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square by substituting the value of $y=fracax+bAx+B$ but failed.
functions
asked Mar 13 at 10:44
Samar Imam ZaidiSamar Imam Zaidi
1,5751520
$endgroup$
Prove that the maximum and minimum values of $y=fracax^2+2bx+cAx^2+2Bx+C$ are those for which $(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square.
I tried to solve this problem using the value of $y'=0$.
Solving the I got $y=fracax+bAx+B$.
Now I need to prove that $T=(ax^2+2bx+c)-y(Ax^2+2Bx+C)$ is a perfect square by substituting the value of $y=fracax+bAx+B$ but failed.
functions
functions
asked Mar 13 at 10:44
Samar Imam ZaidiSamar Imam Zaidi
1,5751520
asked Mar 13 at 10:44
Samar Imam ZaidiSamar Imam Zaidi
1,5751520
asked Mar 13 at 10:44
Samar Imam ZaidiSamar Imam Zaidi
1,5751520
asked Mar 13 at 10:44
Samar Imam ZaidiSamar Imam Zaidi
1,5751520
asked Mar 13 at 10:44
Samar Imam ZaidiSamar Imam Zaidi
1,5751520
1,5751520
$begingroup$
For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
$endgroup$
– Dave L. Renfro
Mar 13 at 11:08
add a comment |
$begingroup$
For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
$endgroup$
– Dave L. Renfro
Mar 13 at 11:08
$begingroup$
For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
$endgroup$
– Dave L. Renfro
Mar 13 at 11:08
$begingroup$
For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
$endgroup$
– Dave L. Renfro
Mar 13 at 11:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: By the quotient rule we get
$$y'=-2,frac Abx^2-Bax^2+Acx-Cax+Bc-Cb left( Ax^2+2,Bx+
C right) ^2
$$
answered Mar 13 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Hint: By the quotient rule we get
$$y'=-2,frac Abx^2-Bax^2+Acx-Cax+Bc-Cb left( Ax^2+2,Bx+
C right) ^2
$$
answered Mar 13 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
$begingroup$
Hint: By the quotient rule we get
$$y'=-2,frac Abx^2-Bax^2+Acx-Cax+Bc-Cb left( Ax^2+2,Bx+
C right) ^2
$$
answered Mar 13 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
$begingroup$
Hint: By the quotient rule we get
$$y'=-2,frac Abx^2-Bax^2+Acx-Cax+Bc-Cb left( Ax^2+2,Bx+
C right) ^2
$$
answered Mar 13 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
Hint: By the quotient rule we get
$$y'=-2,frac Abx^2-Bax^2+Acx-Cax+Bc-Cb left( Ax^2+2,Bx+
C right) ^2
$$
answered Mar 13 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Mar 13 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Mar 13 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Mar 13 at 11:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
77.9k42866
add a comment |
add a comment |
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$begingroup$
For a non-calculus approach, I suspect the method illustrated here can be used. Indeed, you can probably find this in the 1800s literature I allude to (I gave one specific reference).
$endgroup$
– Dave L. Renfro
Mar 13 at 11:08