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How to check if a function is differentiable (example from MIT)
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Here is a problem from MIT, we have to find the values of $a$ and $b$ which make this function differentiable:
For this function to be differentiable, it must
1) Be continuous at $x=1$
2) Have the same derivative values from both sides at $x=1$.
But the solution from MIT says:
Here, we got $14$ from $2x^5+3x^4+4x^2+5x+6$ at $x=1$, so should not it be equal to $ax^2+bx+6$ at $x=1$ instead of $a+b$?
Link of the problem and solution on MIT OpenCoursware
NOTE: Please let me know if such questions are off-topic here and I'll close it, I am fairly new to this website and am not sure how it works.
calculus derivatives continuity
asked Jul 26 at 10:07
Archil Zhvania
1285
add a comment |Â
up vote
1
down vote
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Here is a problem from MIT, we have to find the values of $a$ and $b$ which make this function differentiable:
For this function to be differentiable, it must
1) Be continuous at $x=1$
2) Have the same derivative values from both sides at $x=1$.
But the solution from MIT says:
Here, we got $14$ from $2x^5+3x^4+4x^2+5x+6$ at $x=1$, so should not it be equal to $ax^2+bx+6$ at $x=1$ instead of $a+b$?
Link of the problem and solution on MIT OpenCoursware
NOTE: Please let me know if such questions are off-topic here and I'll close it, I am fairly new to this website and am not sure how it works.
calculus derivatives continuity
asked Jul 26 at 10:07
Archil Zhvania
1285
2
Regarding your note: this question is definitely on-topic for the site. My only (minor) complaint would be that your previous paragraph ("Here, we got 14 ...") is not very clear about your reasoning. (Also, "from MIT" is not very descriptive; do you have a link?)
– Teepeemm
Jul 26 at 14:56
Meta comment: There is Mathematics Meta for discussing about questions (is this on-topic? ...) and help center for information about how to participate on this site (what questions are on-topic, etc.).
– user202729
Jul 26 at 15:06
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here is a problem from MIT, we have to find the values of $a$ and $b$ which make this function differentiable:
For this function to be differentiable, it must
1) Be continuous at $x=1$
2) Have the same derivative values from both sides at $x=1$.
But the solution from MIT says:
Here, we got $14$ from $2x^5+3x^4+4x^2+5x+6$ at $x=1$, so should not it be equal to $ax^2+bx+6$ at $x=1$ instead of $a+b$?
Link of the problem and solution on MIT OpenCoursware
NOTE: Please let me know if such questions are off-topic here and I'll close it, I am fairly new to this website and am not sure how it works.
calculus derivatives continuity
asked Jul 26 at 10:07
Archil Zhvania
1285
Here is a problem from MIT, we have to find the values of $a$ and $b$ which make this function differentiable:
For this function to be differentiable, it must
1) Be continuous at $x=1$
2) Have the same derivative values from both sides at $x=1$.
But the solution from MIT says:
Here, we got $14$ from $2x^5+3x^4+4x^2+5x+6$ at $x=1$, so should not it be equal to $ax^2+bx+6$ at $x=1$ instead of $a+b$?
Link of the problem and solution on MIT OpenCoursware
NOTE: Please let me know if such questions are off-topic here and I'll close it, I am fairly new to this website and am not sure how it works.
calculus derivatives continuity
asked Jul 26 at 10:07
Archil Zhvania
1285
edited Jul 27 at 6:57
asked Jul 26 at 10:07
Archil Zhvania
1285
asked Jul 26 at 10:07
Archil Zhvania
1285
asked Jul 26 at 10:07
Archil Zhvania
1285
1285
2
Regarding your note: this question is definitely on-topic for the site. My only (minor) complaint would be that your previous paragraph ("Here, we got 14 ...") is not very clear about your reasoning. (Also, "from MIT" is not very descriptive; do you have a link?)
– Teepeemm
Jul 26 at 14:56
Meta comment: There is Mathematics Meta for discussing about questions (is this on-topic? ...) and help center for information about how to participate on this site (what questions are on-topic, etc.).
– user202729
Jul 26 at 15:06
add a comment |Â
2
Regarding your note: this question is definitely on-topic for the site. My only (minor) complaint would be that your previous paragraph ("Here, we got 14 ...") is not very clear about your reasoning. (Also, "from MIT" is not very descriptive; do you have a link?)
– Teepeemm
Jul 26 at 14:56
Meta comment: There is Mathematics Meta for discussing about questions (is this on-topic? ...) and help center for information about how to participate on this site (what questions are on-topic, etc.).
– user202729
Jul 26 at 15:06
2
2
Regarding your note: this question is definitely on-topic for the site. My only (minor) complaint would be that your previous paragraph ("Here, we got 14 ...") is not very clear about your reasoning. (Also, "from MIT" is not very descriptive; do you have a link?)
– Teepeemm
Jul 26 at 14:56
Regarding your note: this question is definitely on-topic for the site. My only (minor) complaint would be that your previous paragraph ("Here, we got 14 ...") is not very clear about your reasoning. (Also, "from MIT" is not very descriptive; do you have a link?)
– Teepeemm
Jul 26 at 14:56
Meta comment: There is Mathematics Meta for discussing about questions (is this on-topic? ...) and help center for information about how to participate on this site (what questions are on-topic, etc.).
– user202729
Jul 26 at 15:06
Meta comment: There is Mathematics Meta for discussing about questions (is this on-topic? ...) and help center for information about how to participate on this site (what questions are on-topic, etc.).
– user202729
Jul 26 at 15:06
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
10
down vote
accepted
$2+3+4+5+6=20$.
$f$ is continuous at $1 iff 20=a+b+6 iff a+b=14$.
answered Jul 26 at 10:11
Fred
37.2k1237
add a comment |Â
up vote
4
down vote
Let indicate
$g(x)=ax^2+bx+6$
$h(x)=2x^5+3x^4+4x^2+5x+6$
then we need
- for continuity $$g(1)=h(1) implies a+b=14$$
- for differentiability $$g'(1)=h'(1)implies 2a+b=35$$
answered Jul 26 at 10:13
gimusi
65k73583
No. We have $a+b=14$.
– Fred
Jul 26 at 10:15
@Fred Yes of course I fixed, just a typo! Thanks
– gimusi
Jul 26 at 10:15
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
$2+3+4+5+6=20$.
$f$ is continuous at $1 iff 20=a+b+6 iff a+b=14$.
answered Jul 26 at 10:11
Fred
37.2k1237
add a comment |Â
up vote
10
down vote
accepted
$2+3+4+5+6=20$.
$f$ is continuous at $1 iff 20=a+b+6 iff a+b=14$.
answered Jul 26 at 10:11
Fred
37.2k1237
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
$2+3+4+5+6=20$.
$f$ is continuous at $1 iff 20=a+b+6 iff a+b=14$.
answered Jul 26 at 10:11
Fred
37.2k1237
$2+3+4+5+6=20$.
$f$ is continuous at $1 iff 20=a+b+6 iff a+b=14$.
answered Jul 26 at 10:11
Fred
37.2k1237
answered Jul 26 at 10:11
Fred
37.2k1237
answered Jul 26 at 10:11
Fred
37.2k1237
answered Jul 26 at 10:11
Fred
37.2k1237
37.2k1237
add a comment |Â
add a comment |Â
up vote
4
down vote
Let indicate
$g(x)=ax^2+bx+6$
$h(x)=2x^5+3x^4+4x^2+5x+6$
then we need
- for continuity $$g(1)=h(1) implies a+b=14$$
- for differentiability $$g'(1)=h'(1)implies 2a+b=35$$
answered Jul 26 at 10:13
gimusi
65k73583
No. We have $a+b=14$.
– Fred
Jul 26 at 10:15
@Fred Yes of course I fixed, just a typo! Thanks
– gimusi
Jul 26 at 10:15
add a comment |Â
up vote
4
down vote
Let indicate
$g(x)=ax^2+bx+6$
$h(x)=2x^5+3x^4+4x^2+5x+6$
then we need
- for continuity $$g(1)=h(1) implies a+b=14$$
- for differentiability $$g'(1)=h'(1)implies 2a+b=35$$
answered Jul 26 at 10:13
gimusi
65k73583
No. We have $a+b=14$.
– Fred
Jul 26 at 10:15
@Fred Yes of course I fixed, just a typo! Thanks
– gimusi
Jul 26 at 10:15
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Let indicate
$g(x)=ax^2+bx+6$
$h(x)=2x^5+3x^4+4x^2+5x+6$
then we need
- for continuity $$g(1)=h(1) implies a+b=14$$
- for differentiability $$g'(1)=h'(1)implies 2a+b=35$$
answered Jul 26 at 10:13
gimusi
65k73583
Let indicate
$g(x)=ax^2+bx+6$
$h(x)=2x^5+3x^4+4x^2+5x+6$
then we need
- for continuity $$g(1)=h(1) implies a+b=14$$
- for differentiability $$g'(1)=h'(1)implies 2a+b=35$$
answered Jul 26 at 10:13
gimusi
65k73583
edited Jul 26 at 10:15
answered Jul 26 at 10:13
gimusi
65k73583
answered Jul 26 at 10:13
gimusi
65k73583
answered Jul 26 at 10:13
gimusi
65k73583
65k73583
No. We have $a+b=14$.
– Fred
Jul 26 at 10:15
@Fred Yes of course I fixed, just a typo! Thanks
– gimusi
Jul 26 at 10:15
add a comment |Â
No. We have $a+b=14$.
– Fred
Jul 26 at 10:15
@Fred Yes of course I fixed, just a typo! Thanks
– gimusi
Jul 26 at 10:15
No. We have $a+b=14$.
– Fred
Jul 26 at 10:15
No. We have $a+b=14$.
– Fred
Jul 26 at 10:15
@Fred Yes of course I fixed, just a typo! Thanks
– gimusi
Jul 26 at 10:15
@Fred Yes of course I fixed, just a typo! Thanks
– gimusi
Jul 26 at 10:15
add a comment |Â
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2
Regarding your note: this question is definitely on-topic for the site. My only (minor) complaint would be that your previous paragraph ("Here, we got 14 ...") is not very clear about your reasoning. (Also, "from MIT" is not very descriptive; do you have a link?)
– Teepeemm
Jul 26 at 14:56
Meta comment: There is Mathematics Meta for discussing about questions (is this on-topic? ...) and help center for information about how to participate on this site (what questions are on-topic, etc.).
– user202729
Jul 26 at 15:06