List of critical points

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In TI-Nspire CX CAS, writing:

exp▶list(fMax(sin(x),x,0,10),x)

I get:

((π)/(2)),((5*π)/(2))

while in Wolfram Mathematica, writing:

Maximize[Sin[x], 0 <= x <= 10, x]

I get:

1, x -> π/2

Is there a way to get the result above?

calculus-and-analysis maximum

share|improve this question

edited Aug 18 at 8:16

Carl Woll

55.6k271144

asked Aug 18 at 7:46

TeM

1,648618

add a comment | 

up vote
4
down vote

favorite

1

In TI-Nspire CX CAS, writing:

exp▶list(fMax(sin(x),x,0,10),x)

I get:

((π)/(2)),((5*π)/(2))

while in Wolfram Mathematica, writing:

Maximize[Sin[x], 0 <= x <= 10, x]

I get:

1, x -> π/2

Is there a way to get the result above?

calculus-and-analysis maximum

share|improve this question

edited Aug 18 at 8:16

Carl Woll

55.6k271144

asked Aug 18 at 7:46

TeM

1,648618

add a comment | 

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

In TI-Nspire CX CAS, writing:

exp▶list(fMax(sin(x),x,0,10),x)

I get:

((π)/(2)),((5*π)/(2))

while in Wolfram Mathematica, writing:

Maximize[Sin[x], 0 <= x <= 10, x]

I get:

1, x -> π/2

Is there a way to get the result above?

calculus-and-analysis maximum

share|improve this question

edited Aug 18 at 8:16

Carl Woll

55.6k271144

asked Aug 18 at 7:46

TeM

1,648618

In TI-Nspire CX CAS, writing:

exp▶list(fMax(sin(x),x,0,10),x)

I get:

((π)/(2)),((5*π)/(2))

while in Wolfram Mathematica, writing:

Maximize[Sin[x], 0 <= x <= 10, x]

I get:

1, x -> π/2

Is there a way to get the result above?

calculus-and-analysis maximum

share|improve this question

edited Aug 18 at 8:16

Carl Woll

55.6k271144

asked Aug 18 at 7:46

TeM

1,648618

share|improve this question

share|improve this question

edited Aug 18 at 8:16

Carl Woll

55.6k271144

edited Aug 18 at 8:16

Carl Woll

55.6k271144

edited Aug 18 at 8:16

Carl Woll

55.6k271144

55.6k271144

asked Aug 18 at 7:46

TeM

1,648618

asked Aug 18 at 7:46

TeM

1,648618

asked Aug 18 at 7:46

TeM

1,648618

1,648618

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
4
down vote



accepted

Reduce[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]

x == π/2 || x == (5 π)/2

List @@ %

x == π/2, x == (5 π)/2

Alternatively,

Solve[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]

x -> π/2, x -> (5 π)/2

 x /. %

π/2, (5 π)/2

or

x /. ToRules[Reduce[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]]

π/2, (5 π)/2

share|improve this answer

edited Aug 18 at 8:05

answered Aug 18 at 7:59

kglr

158k8182380

add a comment | 

up vote
6
down vote

I would use MaxValue in combination with Solve:

maxima[expr_, cond_, v_] := With[m = MaxValue[expr, cond, v],
 DeleteDuplicates @ Solve[expr == m && cond, v]
]

For your example:

maxima[Sin[x], 0<=x<=10, x]

x -> π/2, x -> (5 π)/2

Contrast this with the approach in kglr's answer that only looks at critical points:

maxima[Sin[x], -1 <= x <= 1, x]

Reduce[Sin'[x]==0,Sin''[x]<=0,-1<=x<=1,x,Reals]

x -> 1

False

share|improve this answer

answered Aug 18 at 8:15

Carl Woll

55.6k271144

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Reduce[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]

x == π/2 || x == (5 π)/2

List @@ %

x == π/2, x == (5 π)/2

Alternatively,

Solve[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]

x -> π/2, x -> (5 π)/2

 x /. %

π/2, (5 π)/2

or

x /. ToRules[Reduce[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]]

π/2, (5 π)/2

share|improve this answer

edited Aug 18 at 8:05

answered Aug 18 at 7:59

kglr

158k8182380

add a comment | 

up vote
4
down vote



accepted

Reduce[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]

x == π/2 || x == (5 π)/2

List @@ %

x == π/2, x == (5 π)/2

Alternatively,

Solve[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]

x -> π/2, x -> (5 π)/2

 x /. %

π/2, (5 π)/2

or

x /. ToRules[Reduce[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]]

π/2, (5 π)/2

share|improve this answer

edited Aug 18 at 8:05

answered Aug 18 at 7:59

kglr

158k8182380

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Reduce[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]

x == π/2 || x == (5 π)/2

List @@ %

x == π/2, x == (5 π)/2

Alternatively,

Solve[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]

x -> π/2, x -> (5 π)/2

 x /. %

π/2, (5 π)/2

or

x /. ToRules[Reduce[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]]

π/2, (5 π)/2

share|improve this answer

edited Aug 18 at 8:05

answered Aug 18 at 7:59

kglr

158k8182380

Reduce[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]

x == π/2 || x == (5 π)/2

List @@ %

x == π/2, x == (5 π)/2

Alternatively,

Solve[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]

x -> π/2, x -> (5 π)/2

 x /. %

π/2, (5 π)/2

or

x /. ToRules[Reduce[Sin'[x] == 0, Sin''[x] <= 0, 0 <= x <= 10, x, Reals]]

π/2, (5 π)/2

share|improve this answer

edited Aug 18 at 8:05

answered Aug 18 at 7:59

kglr

158k8182380

share|improve this answer

share|improve this answer

edited Aug 18 at 8:05

edited Aug 18 at 8:05

edited Aug 18 at 8:05

answered Aug 18 at 7:59

kglr

158k8182380

answered Aug 18 at 7:59

kglr

158k8182380

answered Aug 18 at 7:59

kglr

158k8182380

158k8182380

add a comment | 

add a comment | 

up vote
6
down vote

I would use MaxValue in combination with Solve:

maxima[expr_, cond_, v_] := With[m = MaxValue[expr, cond, v],
 DeleteDuplicates @ Solve[expr == m && cond, v]
]

For your example:

maxima[Sin[x], 0<=x<=10, x]

x -> π/2, x -> (5 π)/2

Contrast this with the approach in kglr's answer that only looks at critical points:

maxima[Sin[x], -1 <= x <= 1, x]

Reduce[Sin'[x]==0,Sin''[x]<=0,-1<=x<=1,x,Reals]

x -> 1

False

share|improve this answer

answered Aug 18 at 8:15

Carl Woll

55.6k271144

add a comment | 

up vote
6
down vote

I would use MaxValue in combination with Solve:

maxima[expr_, cond_, v_] := With[m = MaxValue[expr, cond, v],
 DeleteDuplicates @ Solve[expr == m && cond, v]
]

For your example:

maxima[Sin[x], 0<=x<=10, x]

x -> π/2, x -> (5 π)/2

Contrast this with the approach in kglr's answer that only looks at critical points:

maxima[Sin[x], -1 <= x <= 1, x]

Reduce[Sin'[x]==0,Sin''[x]<=0,-1<=x<=1,x,Reals]

x -> 1

False

share|improve this answer

answered Aug 18 at 8:15

Carl Woll

55.6k271144

add a comment | 

up vote
6
down vote

up vote
6
down vote

I would use MaxValue in combination with Solve:

maxima[expr_, cond_, v_] := With[m = MaxValue[expr, cond, v],
 DeleteDuplicates @ Solve[expr == m && cond, v]
]

For your example:

maxima[Sin[x], 0<=x<=10, x]

x -> π/2, x -> (5 π)/2

Contrast this with the approach in kglr's answer that only looks at critical points:

maxima[Sin[x], -1 <= x <= 1, x]

Reduce[Sin'[x]==0,Sin''[x]<=0,-1<=x<=1,x,Reals]

x -> 1

False

share|improve this answer

answered Aug 18 at 8:15

Carl Woll

55.6k271144

I would use MaxValue in combination with Solve:

maxima[expr_, cond_, v_] := With[m = MaxValue[expr, cond, v],
 DeleteDuplicates @ Solve[expr == m && cond, v]
]

For your example:

maxima[Sin[x], 0<=x<=10, x]

x -> π/2, x -> (5 π)/2

Contrast this with the approach in kglr's answer that only looks at critical points:

maxima[Sin[x], -1 <= x <= 1, x]

Reduce[Sin'[x]==0,Sin''[x]<=0,-1<=x<=1,x,Reals]

x -> 1

False

share|improve this answer

answered Aug 18 at 8:15

Carl Woll

55.6k271144

share|improve this answer

share|improve this answer

answered Aug 18 at 8:15

Carl Woll

55.6k271144

answered Aug 18 at 8:15

Carl Woll

55.6k271144

answered Aug 18 at 8:15

Carl Woll

55.6k271144

55.6k271144

add a comment | 

add a comment | 

 

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