Difference between the absolute maximum and minimum of $f(x)=2x^3–9x^2+12x+5$
$begingroup$
Let $M$ and $m$ be respectively the absolute maximum and the absolute minimum value of the function, $f(x)=2x^3–9x^2+12x+5$ in the interval $[0,3]$. Then $M–m$ is equal to:
The derivative is $(x-1)(x-2)$ local minimum at 2 is 9 and the local maximum at 1 is 10.However the question asks for the difference between the global maximum and minimum. So I checked the function at 0 and 3 as well which yield 5 and 14 respectively. Using the first set of points at local maximum gives me the answer as 1 but using the latter set of points give me a 9.
This was a question asked in the Joint Engineering Examination thus so far i haven't been able to obtain an official answer key.
Some sites suggest that the answer is 1 while others suggest it as 9. I vaguely recall being taught that if an interval is open then we do not consider the maxima/minima at the end points of an interval but over here the interval is closed so this shouldn't impose a problem. 9 is the correct answer, am I right?
calculus
edited Dec 25 '18 at 16:30
Bernard
121k740116
asked Dec 25 '18 at 16:17
CaptainQuestionCaptainQuestion
1337
$endgroup$
add a comment |
$begingroup$
Let $M$ and $m$ be respectively the absolute maximum and the absolute minimum value of the function, $f(x)=2x^3–9x^2+12x+5$ in the interval $[0,3]$. Then $M–m$ is equal to:
The derivative is $(x-1)(x-2)$ local minimum at 2 is 9 and the local maximum at 1 is 10.However the question asks for the difference between the global maximum and minimum. So I checked the function at 0 and 3 as well which yield 5 and 14 respectively. Using the first set of points at local maximum gives me the answer as 1 but using the latter set of points give me a 9.
This was a question asked in the Joint Engineering Examination thus so far i haven't been able to obtain an official answer key.
Some sites suggest that the answer is 1 while others suggest it as 9. I vaguely recall being taught that if an interval is open then we do not consider the maxima/minima at the end points of an interval but over here the interval is closed so this shouldn't impose a problem. 9 is the correct answer, am I right?
calculus
edited Dec 25 '18 at 16:30
Bernard
121k740116
asked Dec 25 '18 at 16:17
CaptainQuestionCaptainQuestion
1337
$endgroup$
2
$begingroup$
English hint: maxima and minima are plural. The singulars are maximum and minimum.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:24
1
$begingroup$
I think nine is the correct answer since the interval is closed.
$endgroup$
– Larry
Dec 25 '18 at 16:26
add a comment |
$begingroup$
Let $M$ and $m$ be respectively the absolute maximum and the absolute minimum value of the function, $f(x)=2x^3–9x^2+12x+5$ in the interval $[0,3]$. Then $M–m$ is equal to:
The derivative is $(x-1)(x-2)$ local minimum at 2 is 9 and the local maximum at 1 is 10.However the question asks for the difference between the global maximum and minimum. So I checked the function at 0 and 3 as well which yield 5 and 14 respectively. Using the first set of points at local maximum gives me the answer as 1 but using the latter set of points give me a 9.
This was a question asked in the Joint Engineering Examination thus so far i haven't been able to obtain an official answer key.
Some sites suggest that the answer is 1 while others suggest it as 9. I vaguely recall being taught that if an interval is open then we do not consider the maxima/minima at the end points of an interval but over here the interval is closed so this shouldn't impose a problem. 9 is the correct answer, am I right?
calculus
edited Dec 25 '18 at 16:30
Bernard
121k740116
asked Dec 25 '18 at 16:17
CaptainQuestionCaptainQuestion
1337
$endgroup$
Let $M$ and $m$ be respectively the absolute maximum and the absolute minimum value of the function, $f(x)=2x^3–9x^2+12x+5$ in the interval $[0,3]$. Then $M–m$ is equal to:
The derivative is $(x-1)(x-2)$ local minimum at 2 is 9 and the local maximum at 1 is 10.However the question asks for the difference between the global maximum and minimum. So I checked the function at 0 and 3 as well which yield 5 and 14 respectively. Using the first set of points at local maximum gives me the answer as 1 but using the latter set of points give me a 9.
This was a question asked in the Joint Engineering Examination thus so far i haven't been able to obtain an official answer key.
Some sites suggest that the answer is 1 while others suggest it as 9. I vaguely recall being taught that if an interval is open then we do not consider the maxima/minima at the end points of an interval but over here the interval is closed so this shouldn't impose a problem. 9 is the correct answer, am I right?
calculus
calculus
edited Dec 25 '18 at 16:30
Bernard
121k740116
asked Dec 25 '18 at 16:17
CaptainQuestionCaptainQuestion
1337
edited Dec 25 '18 at 16:30
Bernard
121k740116
asked Dec 25 '18 at 16:17
CaptainQuestionCaptainQuestion
1337
edited Dec 25 '18 at 16:30
Bernard
121k740116
edited Dec 25 '18 at 16:30
Bernard
121k740116
edited Dec 25 '18 at 16:30
Bernard
121k740116
121k740116
asked Dec 25 '18 at 16:17
CaptainQuestionCaptainQuestion
1337
asked Dec 25 '18 at 16:17
CaptainQuestionCaptainQuestion
1337
asked Dec 25 '18 at 16:17
CaptainQuestionCaptainQuestion
1337
1337
2
$begingroup$
English hint: maxima and minima are plural. The singulars are maximum and minimum.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:24
1
$begingroup$
I think nine is the correct answer since the interval is closed.
$endgroup$
– Larry
Dec 25 '18 at 16:26
add a comment |
2
$begingroup$
English hint: maxima and minima are plural. The singulars are maximum and minimum.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:24
1
$begingroup$
I think nine is the correct answer since the interval is closed.
$endgroup$
– Larry
Dec 25 '18 at 16:26
2
2
$begingroup$
English hint: maxima and minima are plural. The singulars are maximum and minimum.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:24
$begingroup$
English hint: maxima and minima are plural. The singulars are maximum and minimum.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:24
1
1
$begingroup$
I think nine is the correct answer since the interval is closed.
$endgroup$
– Larry
Dec 25 '18 at 16:26
$begingroup$
I think nine is the correct answer since the interval is closed.
$endgroup$
– Larry
Dec 25 '18 at 16:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your approach is good. You compare the values at the critical points and the endpoints to find the global maximum, which here is $14$. You do the same to find the global minimum, which is $5$. The difference is $9$.
If the interval were open at both ends it would not attain $5$ or $14$ but would get as close as you want. There would not be a global maximum or minimum in that case and the question would not have an answer.
answered Dec 25 '18 at 16:23
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
This not true, sorry. It is $$14-5=9$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 16:25
$begingroup$
@Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:48
add a comment |
$begingroup$
Sometimes a figure is worth a thousand words:
answered Dec 25 '18 at 16:43
David G. StorkDavid G. Stork
11k41432
$endgroup$
add a comment |
$begingroup$
The critical values are the roots of $;f'(x)=6x^2-18x+12=6(x-1)(x-2)$. Furthermore, as the leading coefficient of this cubic polynomial is positive, a local maximum is first attained at $1$, and a local minimum at $2$.
So the global maximum on $[0,3]$ is attained either at $1$ or at $3$. Similarly the global minimum on $[0,3]$ is attained either at $2$ or $0$. To find them we only have to compute the values of $f(x)$ at these points:
$$begin{array}{r!{}cccc}
x=&0&1&2&3 \
hline
f(x)=&5&10&9&10
end{array}$$
We see the global maximum is $M=10$ and the global minimum is $m=5$, so
$$M-m=5.*$$
answered Dec 25 '18 at 17:05
BernardBernard
121k740116
$endgroup$
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
$begingroup$
Your approach is good. You compare the values at the critical points and the endpoints to find the global maximum, which here is $14$. You do the same to find the global minimum, which is $5$. The difference is $9$.
If the interval were open at both ends it would not attain $5$ or $14$ but would get as close as you want. There would not be a global maximum or minimum in that case and the question would not have an answer.
answered Dec 25 '18 at 16:23
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
This not true, sorry. It is $$14-5=9$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 16:25
$begingroup$
@Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:48
add a comment |
$begingroup$
Your approach is good. You compare the values at the critical points and the endpoints to find the global maximum, which here is $14$. You do the same to find the global minimum, which is $5$. The difference is $9$.
If the interval were open at both ends it would not attain $5$ or $14$ but would get as close as you want. There would not be a global maximum or minimum in that case and the question would not have an answer.
answered Dec 25 '18 at 16:23
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
This not true, sorry. It is $$14-5=9$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 16:25
$begingroup$
@Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:48
add a comment |
$begingroup$
Your approach is good. You compare the values at the critical points and the endpoints to find the global maximum, which here is $14$. You do the same to find the global minimum, which is $5$. The difference is $9$.
If the interval were open at both ends it would not attain $5$ or $14$ but would get as close as you want. There would not be a global maximum or minimum in that case and the question would not have an answer.
answered Dec 25 '18 at 16:23
Ross MillikanRoss Millikan
297k23198371
$endgroup$
Your approach is good. You compare the values at the critical points and the endpoints to find the global maximum, which here is $14$. You do the same to find the global minimum, which is $5$. The difference is $9$.
If the interval were open at both ends it would not attain $5$ or $14$ but would get as close as you want. There would not be a global maximum or minimum in that case and the question would not have an answer.
answered Dec 25 '18 at 16:23
Ross MillikanRoss Millikan
297k23198371
answered Dec 25 '18 at 16:23
Ross MillikanRoss Millikan
297k23198371
answered Dec 25 '18 at 16:23
Ross MillikanRoss Millikan
297k23198371
answered Dec 25 '18 at 16:23
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
This not true, sorry. It is $$14-5=9$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 16:25
$begingroup$
@Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:48
add a comment |
$begingroup$
This not true, sorry. It is $$14-5=9$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 16:25
$begingroup$
@Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:48
$begingroup$
This not true, sorry. It is $$14-5=9$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 16:25
$begingroup$
This not true, sorry. It is $$14-5=9$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 25 '18 at 16:25
$begingroup$
@Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:48
$begingroup$
@Dr.SonnhardGraubner: are you referring to the second paragraph? With an open interval there is not $x$ with $f(x)=14$, nor one with $f(x)=5$. The sup of the difference in function values is $9$, but there is no maximum difference in function values. Maxima must be attained.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:48
add a comment |
$begingroup$
Sometimes a figure is worth a thousand words:
answered Dec 25 '18 at 16:43
David G. StorkDavid G. Stork
11k41432
$endgroup$
add a comment |
$begingroup$
Sometimes a figure is worth a thousand words:
answered Dec 25 '18 at 16:43
David G. StorkDavid G. Stork
11k41432
$endgroup$
add a comment |
$begingroup$
Sometimes a figure is worth a thousand words:
answered Dec 25 '18 at 16:43
David G. StorkDavid G. Stork
11k41432
$endgroup$
Sometimes a figure is worth a thousand words:
answered Dec 25 '18 at 16:43
David G. StorkDavid G. Stork
11k41432
answered Dec 25 '18 at 16:43
David G. StorkDavid G. Stork
11k41432
answered Dec 25 '18 at 16:43
David G. StorkDavid G. Stork
11k41432
answered Dec 25 '18 at 16:43
David G. StorkDavid G. Stork
11k41432
11k41432
add a comment |
add a comment |
$begingroup$
The critical values are the roots of $;f'(x)=6x^2-18x+12=6(x-1)(x-2)$. Furthermore, as the leading coefficient of this cubic polynomial is positive, a local maximum is first attained at $1$, and a local minimum at $2$.
So the global maximum on $[0,3]$ is attained either at $1$ or at $3$. Similarly the global minimum on $[0,3]$ is attained either at $2$ or $0$. To find them we only have to compute the values of $f(x)$ at these points:
$$begin{array}{r!{}cccc}
x=&0&1&2&3 \
hline
f(x)=&5&10&9&10
end{array}$$
We see the global maximum is $M=10$ and the global minimum is $m=5$, so
$$M-m=5.*$$
answered Dec 25 '18 at 17:05
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
The critical values are the roots of $;f'(x)=6x^2-18x+12=6(x-1)(x-2)$. Furthermore, as the leading coefficient of this cubic polynomial is positive, a local maximum is first attained at $1$, and a local minimum at $2$.
So the global maximum on $[0,3]$ is attained either at $1$ or at $3$. Similarly the global minimum on $[0,3]$ is attained either at $2$ or $0$. To find them we only have to compute the values of $f(x)$ at these points:
$$begin{array}{r!{}cccc}
x=&0&1&2&3 \
hline
f(x)=&5&10&9&10
end{array}$$
We see the global maximum is $M=10$ and the global minimum is $m=5$, so
$$M-m=5.*$$
answered Dec 25 '18 at 17:05
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
The critical values are the roots of $;f'(x)=6x^2-18x+12=6(x-1)(x-2)$. Furthermore, as the leading coefficient of this cubic polynomial is positive, a local maximum is first attained at $1$, and a local minimum at $2$.
So the global maximum on $[0,3]$ is attained either at $1$ or at $3$. Similarly the global minimum on $[0,3]$ is attained either at $2$ or $0$. To find them we only have to compute the values of $f(x)$ at these points:
$$begin{array}{r!{}cccc}
x=&0&1&2&3 \
hline
f(x)=&5&10&9&10
end{array}$$
We see the global maximum is $M=10$ and the global minimum is $m=5$, so
$$M-m=5.*$$
answered Dec 25 '18 at 17:05
BernardBernard
121k740116
$endgroup$
The critical values are the roots of $;f'(x)=6x^2-18x+12=6(x-1)(x-2)$. Furthermore, as the leading coefficient of this cubic polynomial is positive, a local maximum is first attained at $1$, and a local minimum at $2$.
So the global maximum on $[0,3]$ is attained either at $1$ or at $3$. Similarly the global minimum on $[0,3]$ is attained either at $2$ or $0$. To find them we only have to compute the values of $f(x)$ at these points:
$$begin{array}{r!{}cccc}
x=&0&1&2&3 \
hline
f(x)=&5&10&9&10
end{array}$$
We see the global maximum is $M=10$ and the global minimum is $m=5$, so
$$M-m=5.*$$
answered Dec 25 '18 at 17:05
BernardBernard
121k740116
answered Dec 25 '18 at 17:05
BernardBernard
121k740116
answered Dec 25 '18 at 17:05
BernardBernard
121k740116
answered Dec 25 '18 at 17:05
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
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$begingroup$
English hint: maxima and minima are plural. The singulars are maximum and minimum.
$endgroup$
– Ross Millikan
Dec 25 '18 at 16:24
1
$begingroup$
I think nine is the correct answer since the interval is closed.
$endgroup$
– Larry
Dec 25 '18 at 16:26