How can I find the critical numbers of this function? [closed]
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How can I find the critical numbers of this function?
$$g(theta) = 24theta − 6 tan(theta)$$
calculus
edited Nov 25 at 21:42
NoChance
3,59621221
asked Nov 25 at 21:31
user597553
133
closed as off-topic by caverac, user302797, Leucippus, Chinnapparaj R, user10354138 Nov 26 at 8:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – caverac, user302797, Leucippus, Chinnapparaj R, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-3
down vote
favorite
How can I find the critical numbers of this function?
$$g(theta) = 24theta − 6 tan(theta)$$
calculus
edited Nov 25 at 21:42
NoChance
3,59621221
asked Nov 25 at 21:31
user597553
133
closed as off-topic by caverac, user302797, Leucippus, Chinnapparaj R, user10354138 Nov 26 at 8:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – caverac, user302797, Leucippus, Chinnapparaj R, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
Can you show us what you have attempted? Or what is interesting about this question? Usually questions posed with the tone "do this homework problem for me" are not well-received.
– Mason
Nov 25 at 21:36
"critical numbers" is not a precise term, I think you could say "critical values".
– NoChance
Nov 25 at 21:43
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
How can I find the critical numbers of this function?
$$g(theta) = 24theta − 6 tan(theta)$$
calculus
edited Nov 25 at 21:42
NoChance
3,59621221
asked Nov 25 at 21:31
user597553
133
How can I find the critical numbers of this function?
$$g(theta) = 24theta − 6 tan(theta)$$
calculus
calculus
edited Nov 25 at 21:42
NoChance
3,59621221
asked Nov 25 at 21:31
user597553
133
edited Nov 25 at 21:42
NoChance
3,59621221
asked Nov 25 at 21:31
user597553
133
edited Nov 25 at 21:42
NoChance
3,59621221
edited Nov 25 at 21:42
NoChance
3,59621221
edited Nov 25 at 21:42
NoChance
3,59621221
3,59621221
asked Nov 25 at 21:31
user597553
133
asked Nov 25 at 21:31
user597553
133
asked Nov 25 at 21:31
user597553
133
133
closed as off-topic by caverac, user302797, Leucippus, Chinnapparaj R, user10354138 Nov 26 at 8:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – caverac, user302797, Leucippus, Chinnapparaj R, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by caverac, user302797, Leucippus, Chinnapparaj R, user10354138 Nov 26 at 8:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – caverac, user302797, Leucippus, Chinnapparaj R, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
Can you show us what you have attempted? Or what is interesting about this question? Usually questions posed with the tone "do this homework problem for me" are not well-received.
– Mason
Nov 25 at 21:36
"critical numbers" is not a precise term, I think you could say "critical values".
– NoChance
Nov 25 at 21:43
add a comment |
Can you show us what you have attempted? Or what is interesting about this question? Usually questions posed with the tone "do this homework problem for me" are not well-received.
– Mason
Nov 25 at 21:36
"critical numbers" is not a precise term, I think you could say "critical values".
– NoChance
Nov 25 at 21:43
Can you show us what you have attempted? Or what is interesting about this question? Usually questions posed with the tone "do this homework problem for me" are not well-received.
– Mason
Nov 25 at 21:36
Can you show us what you have attempted? Or what is interesting about this question? Usually questions posed with the tone "do this homework problem for me" are not well-received.
– Mason
Nov 25 at 21:36
"critical numbers" is not a precise term, I think you could say "critical values".
– NoChance
Nov 25 at 21:43
"critical numbers" is not a precise term, I think you could say "critical values".
– NoChance
Nov 25 at 21:43
add a comment |
1 Answer
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Hint: You have $g(theta) = 24theta − 6 tan(theta)$, which implies $$g'(theta) = 24 - 6sec^2(theta) = 0$$ $$implies 24 = 6sec^2(theta)$$ $$implies 4 = {1overcos^2(theta)}$$ $$implies {1over 4} = cos^2(theta)$$ $$implies pm{1over2} = cos(theta).$$ Now, proceed to ask yourself what values of $theta$ will produce this result.
answered Nov 25 at 21:35
Decaf-Math
3,104825
Thank you for your help!
– user597553
Nov 26 at 3:16
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: You have $g(theta) = 24theta − 6 tan(theta)$, which implies $$g'(theta) = 24 - 6sec^2(theta) = 0$$ $$implies 24 = 6sec^2(theta)$$ $$implies 4 = {1overcos^2(theta)}$$ $$implies {1over 4} = cos^2(theta)$$ $$implies pm{1over2} = cos(theta).$$ Now, proceed to ask yourself what values of $theta$ will produce this result.
answered Nov 25 at 21:35
Decaf-Math
3,104825
Thank you for your help!
– user597553
Nov 26 at 3:16
add a comment |
up vote
1
down vote
Hint: You have $g(theta) = 24theta − 6 tan(theta)$, which implies $$g'(theta) = 24 - 6sec^2(theta) = 0$$ $$implies 24 = 6sec^2(theta)$$ $$implies 4 = {1overcos^2(theta)}$$ $$implies {1over 4} = cos^2(theta)$$ $$implies pm{1over2} = cos(theta).$$ Now, proceed to ask yourself what values of $theta$ will produce this result.
answered Nov 25 at 21:35
Decaf-Math
3,104825
Thank you for your help!
– user597553
Nov 26 at 3:16
add a comment |
up vote
1
down vote
up vote
1
down vote
Hint: You have $g(theta) = 24theta − 6 tan(theta)$, which implies $$g'(theta) = 24 - 6sec^2(theta) = 0$$ $$implies 24 = 6sec^2(theta)$$ $$implies 4 = {1overcos^2(theta)}$$ $$implies {1over 4} = cos^2(theta)$$ $$implies pm{1over2} = cos(theta).$$ Now, proceed to ask yourself what values of $theta$ will produce this result.
answered Nov 25 at 21:35
Decaf-Math
3,104825
Hint: You have $g(theta) = 24theta − 6 tan(theta)$, which implies $$g'(theta) = 24 - 6sec^2(theta) = 0$$ $$implies 24 = 6sec^2(theta)$$ $$implies 4 = {1overcos^2(theta)}$$ $$implies {1over 4} = cos^2(theta)$$ $$implies pm{1over2} = cos(theta).$$ Now, proceed to ask yourself what values of $theta$ will produce this result.
answered Nov 25 at 21:35
Decaf-Math
3,104825
answered Nov 25 at 21:35
Decaf-Math
3,104825
answered Nov 25 at 21:35
Decaf-Math
3,104825
answered Nov 25 at 21:35
Decaf-Math
3,104825
3,104825
Thank you for your help!
– user597553
Nov 26 at 3:16
add a comment |
Thank you for your help!
– user597553
Nov 26 at 3:16
Thank you for your help!
– user597553
Nov 26 at 3:16
Thank you for your help!
– user597553
Nov 26 at 3:16
add a comment |
Can you show us what you have attempted? Or what is interesting about this question? Usually questions posed with the tone "do this homework problem for me" are not well-received.
– Mason
Nov 25 at 21:36
"critical numbers" is not a precise term, I think you could say "critical values".
– NoChance
Nov 25 at 21:43