A question on the parabola












0












$begingroup$


$f:[-2,5] to mathbb{R}$



$f(x)= x^2-2mx+m+2$ The smallest value that the function can take is $-10$. Find the sum of the values that $m$ can take.






algebra-precalculus







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$endgroup$












  • $begingroup$
    The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
    $endgroup$
    – rogerl
    Dec 9 '18 at 15:27










  • $begingroup$
    You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
    $endgroup$
    – Arthur
    Dec 9 '18 at 15:28










  • $begingroup$
    @rogerl I tried but it didn't work out.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:33










  • $begingroup$
    @Hatm00 keep in mind the minimum value can also occur at boundary, see
    $endgroup$
    – rsadhvika
    Dec 9 '18 at 15:42












  • $begingroup$
    You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:45
















0












$begingroup$


$f:[-2,5] to mathbb{R}$



$f(x)= x^2-2mx+m+2$ The smallest value that the function can take is $-10$. Find the sum of the values that $m$ can take.






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
    $endgroup$
    – rogerl
    Dec 9 '18 at 15:27










  • $begingroup$
    You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
    $endgroup$
    – Arthur
    Dec 9 '18 at 15:28










  • $begingroup$
    @rogerl I tried but it didn't work out.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:33










  • $begingroup$
    @Hatm00 keep in mind the minimum value can also occur at boundary, see
    $endgroup$
    – rsadhvika
    Dec 9 '18 at 15:42












  • $begingroup$
    You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:45














0












0








0





$begingroup$


$f:[-2,5] to mathbb{R}$



$f(x)= x^2-2mx+m+2$ The smallest value that the function can take is $-10$. Find the sum of the values that $m$ can take.






algebra-precalculus







share|cite|improve this question











$endgroup$




$f:[-2,5] to mathbb{R}$



$f(x)= x^2-2mx+m+2$ The smallest value that the function can take is $-10$. Find the sum of the values that $m$ can take.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 15:45









mm-crj

423213




423213










asked Dec 9 '18 at 15:25









Hatm00Hatm00

276




276












  • $begingroup$
    The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
    $endgroup$
    – rogerl
    Dec 9 '18 at 15:27










  • $begingroup$
    You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
    $endgroup$
    – Arthur
    Dec 9 '18 at 15:28










  • $begingroup$
    @rogerl I tried but it didn't work out.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:33










  • $begingroup$
    @Hatm00 keep in mind the minimum value can also occur at boundary, see
    $endgroup$
    – rsadhvika
    Dec 9 '18 at 15:42












  • $begingroup$
    You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:45


















  • $begingroup$
    The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
    $endgroup$
    – rogerl
    Dec 9 '18 at 15:27










  • $begingroup$
    You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
    $endgroup$
    – Arthur
    Dec 9 '18 at 15:28










  • $begingroup$
    @rogerl I tried but it didn't work out.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:33










  • $begingroup$
    @Hatm00 keep in mind the minimum value can also occur at boundary, see
    $endgroup$
    – rsadhvika
    Dec 9 '18 at 15:42












  • $begingroup$
    You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:45
















$begingroup$
The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
$endgroup$
– rogerl
Dec 9 '18 at 15:27




$begingroup$
The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
$endgroup$
– rogerl
Dec 9 '18 at 15:27












$begingroup$
You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
$endgroup$
– Arthur
Dec 9 '18 at 15:28




$begingroup$
You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
$endgroup$
– Arthur
Dec 9 '18 at 15:28












$begingroup$
@rogerl I tried but it didn't work out.
$endgroup$
– Hatm00
Dec 9 '18 at 15:33




$begingroup$
@rogerl I tried but it didn't work out.
$endgroup$
– Hatm00
Dec 9 '18 at 15:33












$begingroup$
@Hatm00 keep in mind the minimum value can also occur at boundary, see
$endgroup$
– rsadhvika
Dec 9 '18 at 15:42






$begingroup$
@Hatm00 keep in mind the minimum value can also occur at boundary, see
$endgroup$
– rsadhvika
Dec 9 '18 at 15:42














$begingroup$
You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
$endgroup$
– Hatm00
Dec 9 '18 at 15:45




$begingroup$
You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
$endgroup$
– Hatm00
Dec 9 '18 at 15:45










2 Answers
2






active

oldest

votes


















2












$begingroup$

The smallest value of $ax^2+bx+c$ (when $a>0$) is taken at $x$ halfway between the roots, which is
$$
x=-frac{b}{2a}=m
$$
(For $a<0$, that abscissa gives the maximum value.)



However, your problem is a bit more complicated, because you bound the function over $[-2,5]$. Then the minimum value can be at $m$ (if it is in this range), at $-2$ or at $5$.



Now you have $f(-2)=5m+6$ and $f(5)=-9m+27$. Also
$$
f(m)=m^2-2m^2+m+2=-m^2+m+2
$$




  • If $-2<m<5$, the minimum value is taken at $m$

  • if $mle-2$, the minimum is taken at …

  • if $mge5$, the minimum is taken at …


We have $f(m)=-10$ if and only if $m^2-m-12=0$, that is $m=4$ or $m=-3$.



We have $f(-2)=-10$ if and only if $m=-16/5<-2$.



We have $f(5)=-10$ if and only if $m=37/9$.



Two of these values have to be discarded.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
    $endgroup$
    – gimusi
    Dec 9 '18 at 15:32












  • $begingroup$
    Hey smallest value can also occur at boundary right ?
    $endgroup$
    – rsadhvika
    Dec 9 '18 at 15:34










  • $begingroup$
    But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:42










  • $begingroup$
    @gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
    $endgroup$
    – egreg
    Dec 9 '18 at 16:04










  • $begingroup$
    @Hatm00 I added further hints
    $endgroup$
    – egreg
    Dec 9 '18 at 16:19



















0












$begingroup$

The maximum or minimum of a parabola occurs at its vertex at $(h, k)$.



$$h = frac{-b}{2a}$$



$$k = c-frac{b^2}{4a}$$



Since you have $a = 1 > 0$, then a minimum exists.



$$k = -10 iff c-frac{b^2}{4a} = -10 iff (m+2)-frac{4m^2}{4} = -10$$



$$frac{4m+8-4m^2}{4} = -10 iff -4m^2+4m+8 = -40$$



$$-4m^2+4m+48 = 0 iff m^2-m-12 = 0 implies (m-4)(m+3) = 0$$



This yields $m = 4$ and $m = -3$, but only the former is applicable given the limited domain.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But we don't know is vertex in definition range.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:39










  • $begingroup$
    @Hatm00 I’ve edited the answer. (I misread the question.)
    $endgroup$
    – KM101
    Dec 9 '18 at 15:51












  • $begingroup$
    Thanks so much for your help. If we add root with the boundary values we can reach the result.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 16:01












  • $begingroup$
    It would be good if the downvoter clarifies if the answer is missing anything.
    $endgroup$
    – KM101
    Dec 9 '18 at 16:03













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

The smallest value of $ax^2+bx+c$ (when $a>0$) is taken at $x$ halfway between the roots, which is
$$
x=-frac{b}{2a}=m
$$
(For $a<0$, that abscissa gives the maximum value.)



However, your problem is a bit more complicated, because you bound the function over $[-2,5]$. Then the minimum value can be at $m$ (if it is in this range), at $-2$ or at $5$.



Now you have $f(-2)=5m+6$ and $f(5)=-9m+27$. Also
$$
f(m)=m^2-2m^2+m+2=-m^2+m+2
$$




  • If $-2<m<5$, the minimum value is taken at $m$

  • if $mle-2$, the minimum is taken at …

  • if $mge5$, the minimum is taken at …


We have $f(m)=-10$ if and only if $m^2-m-12=0$, that is $m=4$ or $m=-3$.



We have $f(-2)=-10$ if and only if $m=-16/5<-2$.



We have $f(5)=-10$ if and only if $m=37/9$.



Two of these values have to be discarded.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
    $endgroup$
    – gimusi
    Dec 9 '18 at 15:32












  • $begingroup$
    Hey smallest value can also occur at boundary right ?
    $endgroup$
    – rsadhvika
    Dec 9 '18 at 15:34










  • $begingroup$
    But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:42










  • $begingroup$
    @gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
    $endgroup$
    – egreg
    Dec 9 '18 at 16:04










  • $begingroup$
    @Hatm00 I added further hints
    $endgroup$
    – egreg
    Dec 9 '18 at 16:19
















2












$begingroup$

The smallest value of $ax^2+bx+c$ (when $a>0$) is taken at $x$ halfway between the roots, which is
$$
x=-frac{b}{2a}=m
$$
(For $a<0$, that abscissa gives the maximum value.)



However, your problem is a bit more complicated, because you bound the function over $[-2,5]$. Then the minimum value can be at $m$ (if it is in this range), at $-2$ or at $5$.



Now you have $f(-2)=5m+6$ and $f(5)=-9m+27$. Also
$$
f(m)=m^2-2m^2+m+2=-m^2+m+2
$$




  • If $-2<m<5$, the minimum value is taken at $m$

  • if $mle-2$, the minimum is taken at …

  • if $mge5$, the minimum is taken at …


We have $f(m)=-10$ if and only if $m^2-m-12=0$, that is $m=4$ or $m=-3$.



We have $f(-2)=-10$ if and only if $m=-16/5<-2$.



We have $f(5)=-10$ if and only if $m=37/9$.



Two of these values have to be discarded.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
    $endgroup$
    – gimusi
    Dec 9 '18 at 15:32












  • $begingroup$
    Hey smallest value can also occur at boundary right ?
    $endgroup$
    – rsadhvika
    Dec 9 '18 at 15:34










  • $begingroup$
    But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:42










  • $begingroup$
    @gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
    $endgroup$
    – egreg
    Dec 9 '18 at 16:04










  • $begingroup$
    @Hatm00 I added further hints
    $endgroup$
    – egreg
    Dec 9 '18 at 16:19














2












2








2





$begingroup$

The smallest value of $ax^2+bx+c$ (when $a>0$) is taken at $x$ halfway between the roots, which is
$$
x=-frac{b}{2a}=m
$$
(For $a<0$, that abscissa gives the maximum value.)



However, your problem is a bit more complicated, because you bound the function over $[-2,5]$. Then the minimum value can be at $m$ (if it is in this range), at $-2$ or at $5$.



Now you have $f(-2)=5m+6$ and $f(5)=-9m+27$. Also
$$
f(m)=m^2-2m^2+m+2=-m^2+m+2
$$




  • If $-2<m<5$, the minimum value is taken at $m$

  • if $mle-2$, the minimum is taken at …

  • if $mge5$, the minimum is taken at …


We have $f(m)=-10$ if and only if $m^2-m-12=0$, that is $m=4$ or $m=-3$.



We have $f(-2)=-10$ if and only if $m=-16/5<-2$.



We have $f(5)=-10$ if and only if $m=37/9$.



Two of these values have to be discarded.






share|cite|improve this answer











$endgroup$



The smallest value of $ax^2+bx+c$ (when $a>0$) is taken at $x$ halfway between the roots, which is
$$
x=-frac{b}{2a}=m
$$
(For $a<0$, that abscissa gives the maximum value.)



However, your problem is a bit more complicated, because you bound the function over $[-2,5]$. Then the minimum value can be at $m$ (if it is in this range), at $-2$ or at $5$.



Now you have $f(-2)=5m+6$ and $f(5)=-9m+27$. Also
$$
f(m)=m^2-2m^2+m+2=-m^2+m+2
$$




  • If $-2<m<5$, the minimum value is taken at $m$

  • if $mle-2$, the minimum is taken at …

  • if $mge5$, the minimum is taken at …


We have $f(m)=-10$ if and only if $m^2-m-12=0$, that is $m=4$ or $m=-3$.



We have $f(-2)=-10$ if and only if $m=-16/5<-2$.



We have $f(5)=-10$ if and only if $m=37/9$.



Two of these values have to be discarded.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 16:19

























answered Dec 9 '18 at 15:28









egregegreg

184k1486206




184k1486206












  • $begingroup$
    When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
    $endgroup$
    – gimusi
    Dec 9 '18 at 15:32












  • $begingroup$
    Hey smallest value can also occur at boundary right ?
    $endgroup$
    – rsadhvika
    Dec 9 '18 at 15:34










  • $begingroup$
    But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:42










  • $begingroup$
    @gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
    $endgroup$
    – egreg
    Dec 9 '18 at 16:04










  • $begingroup$
    @Hatm00 I added further hints
    $endgroup$
    – egreg
    Dec 9 '18 at 16:19


















  • $begingroup$
    When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
    $endgroup$
    – gimusi
    Dec 9 '18 at 15:32












  • $begingroup$
    Hey smallest value can also occur at boundary right ?
    $endgroup$
    – rsadhvika
    Dec 9 '18 at 15:34










  • $begingroup$
    But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:42










  • $begingroup$
    @gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
    $endgroup$
    – egreg
    Dec 9 '18 at 16:04










  • $begingroup$
    @Hatm00 I added further hints
    $endgroup$
    – egreg
    Dec 9 '18 at 16:19
















$begingroup$
When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
$endgroup$
– gimusi
Dec 9 '18 at 15:32






$begingroup$
When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
$endgroup$
– gimusi
Dec 9 '18 at 15:32














$begingroup$
Hey smallest value can also occur at boundary right ?
$endgroup$
– rsadhvika
Dec 9 '18 at 15:34




$begingroup$
Hey smallest value can also occur at boundary right ?
$endgroup$
– rsadhvika
Dec 9 '18 at 15:34












$begingroup$
But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
$endgroup$
– Hatm00
Dec 9 '18 at 15:42




$begingroup$
But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
$endgroup$
– Hatm00
Dec 9 '18 at 15:42












$begingroup$
@gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
$endgroup$
– egreg
Dec 9 '18 at 16:04




$begingroup$
@gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
$endgroup$
– egreg
Dec 9 '18 at 16:04












$begingroup$
@Hatm00 I added further hints
$endgroup$
– egreg
Dec 9 '18 at 16:19




$begingroup$
@Hatm00 I added further hints
$endgroup$
– egreg
Dec 9 '18 at 16:19











0












$begingroup$

The maximum or minimum of a parabola occurs at its vertex at $(h, k)$.



$$h = frac{-b}{2a}$$



$$k = c-frac{b^2}{4a}$$



Since you have $a = 1 > 0$, then a minimum exists.



$$k = -10 iff c-frac{b^2}{4a} = -10 iff (m+2)-frac{4m^2}{4} = -10$$



$$frac{4m+8-4m^2}{4} = -10 iff -4m^2+4m+8 = -40$$



$$-4m^2+4m+48 = 0 iff m^2-m-12 = 0 implies (m-4)(m+3) = 0$$



This yields $m = 4$ and $m = -3$, but only the former is applicable given the limited domain.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But we don't know is vertex in definition range.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:39










  • $begingroup$
    @Hatm00 I’ve edited the answer. (I misread the question.)
    $endgroup$
    – KM101
    Dec 9 '18 at 15:51












  • $begingroup$
    Thanks so much for your help. If we add root with the boundary values we can reach the result.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 16:01












  • $begingroup$
    It would be good if the downvoter clarifies if the answer is missing anything.
    $endgroup$
    – KM101
    Dec 9 '18 at 16:03


















0












$begingroup$

The maximum or minimum of a parabola occurs at its vertex at $(h, k)$.



$$h = frac{-b}{2a}$$



$$k = c-frac{b^2}{4a}$$



Since you have $a = 1 > 0$, then a minimum exists.



$$k = -10 iff c-frac{b^2}{4a} = -10 iff (m+2)-frac{4m^2}{4} = -10$$



$$frac{4m+8-4m^2}{4} = -10 iff -4m^2+4m+8 = -40$$



$$-4m^2+4m+48 = 0 iff m^2-m-12 = 0 implies (m-4)(m+3) = 0$$



This yields $m = 4$ and $m = -3$, but only the former is applicable given the limited domain.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But we don't know is vertex in definition range.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:39










  • $begingroup$
    @Hatm00 I’ve edited the answer. (I misread the question.)
    $endgroup$
    – KM101
    Dec 9 '18 at 15:51












  • $begingroup$
    Thanks so much for your help. If we add root with the boundary values we can reach the result.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 16:01












  • $begingroup$
    It would be good if the downvoter clarifies if the answer is missing anything.
    $endgroup$
    – KM101
    Dec 9 '18 at 16:03
















0












0








0





$begingroup$

The maximum or minimum of a parabola occurs at its vertex at $(h, k)$.



$$h = frac{-b}{2a}$$



$$k = c-frac{b^2}{4a}$$



Since you have $a = 1 > 0$, then a minimum exists.



$$k = -10 iff c-frac{b^2}{4a} = -10 iff (m+2)-frac{4m^2}{4} = -10$$



$$frac{4m+8-4m^2}{4} = -10 iff -4m^2+4m+8 = -40$$



$$-4m^2+4m+48 = 0 iff m^2-m-12 = 0 implies (m-4)(m+3) = 0$$



This yields $m = 4$ and $m = -3$, but only the former is applicable given the limited domain.






share|cite|improve this answer











$endgroup$



The maximum or minimum of a parabola occurs at its vertex at $(h, k)$.



$$h = frac{-b}{2a}$$



$$k = c-frac{b^2}{4a}$$



Since you have $a = 1 > 0$, then a minimum exists.



$$k = -10 iff c-frac{b^2}{4a} = -10 iff (m+2)-frac{4m^2}{4} = -10$$



$$frac{4m+8-4m^2}{4} = -10 iff -4m^2+4m+8 = -40$$



$$-4m^2+4m+48 = 0 iff m^2-m-12 = 0 implies (m-4)(m+3) = 0$$



This yields $m = 4$ and $m = -3$, but only the former is applicable given the limited domain.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 15:47

























answered Dec 9 '18 at 15:35









KM101KM101

6,0901525




6,0901525












  • $begingroup$
    But we don't know is vertex in definition range.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:39










  • $begingroup$
    @Hatm00 I’ve edited the answer. (I misread the question.)
    $endgroup$
    – KM101
    Dec 9 '18 at 15:51












  • $begingroup$
    Thanks so much for your help. If we add root with the boundary values we can reach the result.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 16:01












  • $begingroup$
    It would be good if the downvoter clarifies if the answer is missing anything.
    $endgroup$
    – KM101
    Dec 9 '18 at 16:03




















  • $begingroup$
    But we don't know is vertex in definition range.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 15:39










  • $begingroup$
    @Hatm00 I’ve edited the answer. (I misread the question.)
    $endgroup$
    – KM101
    Dec 9 '18 at 15:51












  • $begingroup$
    Thanks so much for your help. If we add root with the boundary values we can reach the result.
    $endgroup$
    – Hatm00
    Dec 9 '18 at 16:01












  • $begingroup$
    It would be good if the downvoter clarifies if the answer is missing anything.
    $endgroup$
    – KM101
    Dec 9 '18 at 16:03


















$begingroup$
But we don't know is vertex in definition range.
$endgroup$
– Hatm00
Dec 9 '18 at 15:39




$begingroup$
But we don't know is vertex in definition range.
$endgroup$
– Hatm00
Dec 9 '18 at 15:39












$begingroup$
@Hatm00 I’ve edited the answer. (I misread the question.)
$endgroup$
– KM101
Dec 9 '18 at 15:51






$begingroup$
@Hatm00 I’ve edited the answer. (I misread the question.)
$endgroup$
– KM101
Dec 9 '18 at 15:51














$begingroup$
Thanks so much for your help. If we add root with the boundary values we can reach the result.
$endgroup$
– Hatm00
Dec 9 '18 at 16:01






$begingroup$
Thanks so much for your help. If we add root with the boundary values we can reach the result.
$endgroup$
– Hatm00
Dec 9 '18 at 16:01














$begingroup$
It would be good if the downvoter clarifies if the answer is missing anything.
$endgroup$
– KM101
Dec 9 '18 at 16:03






$begingroup$
It would be good if the downvoter clarifies if the answer is missing anything.
$endgroup$
– KM101
Dec 9 '18 at 16:03




















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