Mixing
Maximum vertical distance for concave function.
$begingroup$
I need to find the maximum vertical distance between a parabola $f(x)=3-x^2$ and a line $g(x)=x+1$ on the interval of $[-2,1]$. Usually I see people doing $g(x)-f(x)$, in other words subtracting the parabola from the line. But since in this case we have a concave function does it mean I have to do $f(x)-g(x)$? Or is it always "line - parabola"?
optimization quadratics
edited Jan 27 at 14:41
Michael Rozenberg
109k1896200
asked Jan 27 at 14:01
Nick202Nick202
605
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add a comment |
$begingroup$
I need to find the maximum vertical distance between a parabola $f(x)=3-x^2$ and a line $g(x)=x+1$ on the interval of $[-2,1]$. Usually I see people doing $g(x)-f(x)$, in other words subtracting the parabola from the line. But since in this case we have a concave function does it mean I have to do $f(x)-g(x)$? Or is it always "line - parabola"?
optimization quadratics
edited Jan 27 at 14:41
Michael Rozenberg
109k1896200
asked Jan 27 at 14:01
Nick202Nick202
605
$endgroup$
add a comment |
$begingroup$
I need to find the maximum vertical distance between a parabola $f(x)=3-x^2$ and a line $g(x)=x+1$ on the interval of $[-2,1]$. Usually I see people doing $g(x)-f(x)$, in other words subtracting the parabola from the line. But since in this case we have a concave function does it mean I have to do $f(x)-g(x)$? Or is it always "line - parabola"?
optimization quadratics
edited Jan 27 at 14:41
Michael Rozenberg
109k1896200
asked Jan 27 at 14:01
Nick202Nick202
605
$endgroup$
I need to find the maximum vertical distance between a parabola $f(x)=3-x^2$ and a line $g(x)=x+1$ on the interval of $[-2,1]$. Usually I see people doing $g(x)-f(x)$, in other words subtracting the parabola from the line. But since in this case we have a concave function does it mean I have to do $f(x)-g(x)$? Or is it always "line - parabola"?
optimization quadratics
optimization quadratics
edited Jan 27 at 14:41
Michael Rozenberg
109k1896200
asked Jan 27 at 14:01
Nick202Nick202
605
edited Jan 27 at 14:41
Michael Rozenberg
109k1896200
asked Jan 27 at 14:01
Nick202Nick202
605
edited Jan 27 at 14:41
Michael Rozenberg
109k1896200
edited Jan 27 at 14:41
Michael Rozenberg
109k1896200
edited Jan 27 at 14:41
Michael Rozenberg
109k1896200
109k1896200
asked Jan 27 at 14:01
Nick202Nick202
605
asked Jan 27 at 14:01
Nick202Nick202
605
asked Jan 27 at 14:01
Nick202Nick202
605
605
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
$f(x)geq g(x)$ for all $-2leq xleq1$ and
$$f(x)-g(x)=3-x^2-x-1=frac{9}{4}-frac{1}{4}-x-x^2=frac{9}{4}-left(frac{1}{2}+xright)^2leqfrac{9}{4}.$$
The equality occurs for $x=-frac{1}{2}in[-2,1]$,
which says that $frac{9}{4}$ is the answer.
answered Jan 27 at 14:19
Michael RozenbergMichael Rozenberg
109k1896200
$endgroup$
$begingroup$
Where did you get the $9/4$ and the $1/4$ from?
$endgroup$
– Nick202
Jan 27 at 14:21
$begingroup$
@Nick202 Just $frac{9}{4}-frac{1}{4}=2=3-1.$
$endgroup$
– Michael Rozenberg
Jan 27 at 14:22
add a comment |
$begingroup$
It does not matter. You should be thinking absolute distance: $$|f(x)-g(x)|=|g(x)-f(x)|$$
If you ignore the absolute value, you should be thinking about extremum instead of minimum or maximum, and then you are fine.
answered Jan 27 at 14:13
AndreiAndrei
13.2k21230
$endgroup$
$begingroup$
So I will get a composite function of $|-x^2-x+4|$?
$endgroup$
– Nick202
Jan 27 at 14:15
$begingroup$
$3-x^2-(x+1)=-x^2-x+2$. But yes.
$endgroup$
– Andrei
Jan 27 at 14:17
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$f(x)geq g(x)$ for all $-2leq xleq1$ and
$$f(x)-g(x)=3-x^2-x-1=frac{9}{4}-frac{1}{4}-x-x^2=frac{9}{4}-left(frac{1}{2}+xright)^2leqfrac{9}{4}.$$
The equality occurs for $x=-frac{1}{2}in[-2,1]$,
which says that $frac{9}{4}$ is the answer.
answered Jan 27 at 14:19
Michael RozenbergMichael Rozenberg
109k1896200
$endgroup$
$begingroup$
Where did you get the $9/4$ and the $1/4$ from?
$endgroup$
– Nick202
Jan 27 at 14:21
$begingroup$
@Nick202 Just $frac{9}{4}-frac{1}{4}=2=3-1.$
$endgroup$
– Michael Rozenberg
Jan 27 at 14:22
add a comment |
$begingroup$
$f(x)geq g(x)$ for all $-2leq xleq1$ and
$$f(x)-g(x)=3-x^2-x-1=frac{9}{4}-frac{1}{4}-x-x^2=frac{9}{4}-left(frac{1}{2}+xright)^2leqfrac{9}{4}.$$
The equality occurs for $x=-frac{1}{2}in[-2,1]$,
which says that $frac{9}{4}$ is the answer.
answered Jan 27 at 14:19
Michael RozenbergMichael Rozenberg
109k1896200
$endgroup$
$begingroup$
Where did you get the $9/4$ and the $1/4$ from?
$endgroup$
– Nick202
Jan 27 at 14:21
$begingroup$
@Nick202 Just $frac{9}{4}-frac{1}{4}=2=3-1.$
$endgroup$
– Michael Rozenberg
Jan 27 at 14:22
add a comment |
$begingroup$
$f(x)geq g(x)$ for all $-2leq xleq1$ and
$$f(x)-g(x)=3-x^2-x-1=frac{9}{4}-frac{1}{4}-x-x^2=frac{9}{4}-left(frac{1}{2}+xright)^2leqfrac{9}{4}.$$
The equality occurs for $x=-frac{1}{2}in[-2,1]$,
which says that $frac{9}{4}$ is the answer.
answered Jan 27 at 14:19
Michael RozenbergMichael Rozenberg
109k1896200
$endgroup$
$f(x)geq g(x)$ for all $-2leq xleq1$ and
$$f(x)-g(x)=3-x^2-x-1=frac{9}{4}-frac{1}{4}-x-x^2=frac{9}{4}-left(frac{1}{2}+xright)^2leqfrac{9}{4}.$$
The equality occurs for $x=-frac{1}{2}in[-2,1]$,
which says that $frac{9}{4}$ is the answer.
answered Jan 27 at 14:19
Michael RozenbergMichael Rozenberg
109k1896200
answered Jan 27 at 14:19
Michael RozenbergMichael Rozenberg
109k1896200
answered Jan 27 at 14:19
Michael RozenbergMichael Rozenberg
109k1896200
answered Jan 27 at 14:19
Michael RozenbergMichael Rozenberg
109k1896200
109k1896200
$begingroup$
Where did you get the $9/4$ and the $1/4$ from?
$endgroup$
– Nick202
Jan 27 at 14:21
$begingroup$
@Nick202 Just $frac{9}{4}-frac{1}{4}=2=3-1.$
$endgroup$
– Michael Rozenberg
Jan 27 at 14:22
add a comment |
$begingroup$
Where did you get the $9/4$ and the $1/4$ from?
$endgroup$
– Nick202
Jan 27 at 14:21
$begingroup$
@Nick202 Just $frac{9}{4}-frac{1}{4}=2=3-1.$
$endgroup$
– Michael Rozenberg
Jan 27 at 14:22
$begingroup$
Where did you get the $9/4$ and the $1/4$ from?
$endgroup$
– Nick202
Jan 27 at 14:21
$begingroup$
Where did you get the $9/4$ and the $1/4$ from?
$endgroup$
– Nick202
Jan 27 at 14:21
$begingroup$
@Nick202 Just $frac{9}{4}-frac{1}{4}=2=3-1.$
$endgroup$
– Michael Rozenberg
Jan 27 at 14:22
$begingroup$
@Nick202 Just $frac{9}{4}-frac{1}{4}=2=3-1.$
$endgroup$
– Michael Rozenberg
Jan 27 at 14:22
add a comment |
$begingroup$
It does not matter. You should be thinking absolute distance: $$|f(x)-g(x)|=|g(x)-f(x)|$$
If you ignore the absolute value, you should be thinking about extremum instead of minimum or maximum, and then you are fine.
answered Jan 27 at 14:13
AndreiAndrei
13.2k21230
$endgroup$
$begingroup$
So I will get a composite function of $|-x^2-x+4|$?
$endgroup$
– Nick202
Jan 27 at 14:15
$begingroup$
$3-x^2-(x+1)=-x^2-x+2$. But yes.
$endgroup$
– Andrei
Jan 27 at 14:17
add a comment |
$begingroup$
It does not matter. You should be thinking absolute distance: $$|f(x)-g(x)|=|g(x)-f(x)|$$
If you ignore the absolute value, you should be thinking about extremum instead of minimum or maximum, and then you are fine.
answered Jan 27 at 14:13
AndreiAndrei
13.2k21230
$endgroup$
$begingroup$
So I will get a composite function of $|-x^2-x+4|$?
$endgroup$
– Nick202
Jan 27 at 14:15
$begingroup$
$3-x^2-(x+1)=-x^2-x+2$. But yes.
$endgroup$
– Andrei
Jan 27 at 14:17
add a comment |
$begingroup$
It does not matter. You should be thinking absolute distance: $$|f(x)-g(x)|=|g(x)-f(x)|$$
If you ignore the absolute value, you should be thinking about extremum instead of minimum or maximum, and then you are fine.
answered Jan 27 at 14:13
AndreiAndrei
13.2k21230
$endgroup$
It does not matter. You should be thinking absolute distance: $$|f(x)-g(x)|=|g(x)-f(x)|$$
If you ignore the absolute value, you should be thinking about extremum instead of minimum or maximum, and then you are fine.
answered Jan 27 at 14:13
AndreiAndrei
13.2k21230
answered Jan 27 at 14:13
AndreiAndrei
13.2k21230
answered Jan 27 at 14:13
AndreiAndrei
13.2k21230
answered Jan 27 at 14:13
AndreiAndrei
13.2k21230
13.2k21230
$begingroup$
So I will get a composite function of $|-x^2-x+4|$?
$endgroup$
– Nick202
Jan 27 at 14:15
$begingroup$
$3-x^2-(x+1)=-x^2-x+2$. But yes.
$endgroup$
– Andrei
Jan 27 at 14:17
add a comment |
$begingroup$
So I will get a composite function of $|-x^2-x+4|$?
$endgroup$
– Nick202
Jan 27 at 14:15
$begingroup$
$3-x^2-(x+1)=-x^2-x+2$. But yes.
$endgroup$
– Andrei
Jan 27 at 14:17
$begingroup$
So I will get a composite function of $|-x^2-x+4|$?
$endgroup$
– Nick202
Jan 27 at 14:15
$begingroup$
So I will get a composite function of $|-x^2-x+4|$?
$endgroup$
– Nick202
Jan 27 at 14:15
$begingroup$
$3-x^2-(x+1)=-x^2-x+2$. But yes.
$endgroup$
– Andrei
Jan 27 at 14:17
$begingroup$
$3-x^2-(x+1)=-x^2-x+2$. But yes.
$endgroup$
– Andrei
Jan 27 at 14:17
add a comment |
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