Find the coordinates for the absolute maximum and minimum values of the function on the given interval
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I am trying to learn how to find the coordinates for the absolute maximum and minimum values of the function on the given interval.
$$f(t) = 2-|t|, -1 ≤ t ≤ 3$$
The answer in my textbook says the minimum is $(3, -1)$ and the maximum is $(0, 2)$.
I keep getting different results, and I don't understand what I'm doing wrong. If someone could provide me with the general concept of how to solve these types of questions, I would really appreciate it.
My solution:
$f(-1) = 2-|-1|$
$f(-1) = 1$
$f(3) = 2 - |3|$
$f(3) = -1$
I thought doing that would give me the $y$ values for the absolute max and min, but clearly, it differs from the answer provided in the textbook.
Additionally, I don't really understand how to get the corresponding $x$ values for the max and min. I tried taking the derivative of the original function which left me with $f'(t) = -t/|t|$ but I didn't know where to go from there.
calculus derivatives absolute-value
edited Nov 27 at 6:59
Eevee Trainer
2,466220
asked Nov 27 at 6:46
Blackthorn
174
add a comment |
up vote
0
down vote
favorite
I am trying to learn how to find the coordinates for the absolute maximum and minimum values of the function on the given interval.
$$f(t) = 2-|t|, -1 ≤ t ≤ 3$$
The answer in my textbook says the minimum is $(3, -1)$ and the maximum is $(0, 2)$.
I keep getting different results, and I don't understand what I'm doing wrong. If someone could provide me with the general concept of how to solve these types of questions, I would really appreciate it.
My solution:
$f(-1) = 2-|-1|$
$f(-1) = 1$
$f(3) = 2 - |3|$
$f(3) = -1$
I thought doing that would give me the $y$ values for the absolute max and min, but clearly, it differs from the answer provided in the textbook.
Additionally, I don't really understand how to get the corresponding $x$ values for the max and min. I tried taking the derivative of the original function which left me with $f'(t) = -t/|t|$ but I didn't know where to go from there.
calculus derivatives absolute-value
edited Nov 27 at 6:59
Eevee Trainer
2,466220
asked Nov 27 at 6:46
Blackthorn
174
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to learn how to find the coordinates for the absolute maximum and minimum values of the function on the given interval.
$$f(t) = 2-|t|, -1 ≤ t ≤ 3$$
The answer in my textbook says the minimum is $(3, -1)$ and the maximum is $(0, 2)$.
I keep getting different results, and I don't understand what I'm doing wrong. If someone could provide me with the general concept of how to solve these types of questions, I would really appreciate it.
My solution:
$f(-1) = 2-|-1|$
$f(-1) = 1$
$f(3) = 2 - |3|$
$f(3) = -1$
I thought doing that would give me the $y$ values for the absolute max and min, but clearly, it differs from the answer provided in the textbook.
Additionally, I don't really understand how to get the corresponding $x$ values for the max and min. I tried taking the derivative of the original function which left me with $f'(t) = -t/|t|$ but I didn't know where to go from there.
calculus derivatives absolute-value
edited Nov 27 at 6:59
Eevee Trainer
2,466220
asked Nov 27 at 6:46
Blackthorn
174
I am trying to learn how to find the coordinates for the absolute maximum and minimum values of the function on the given interval.
$$f(t) = 2-|t|, -1 ≤ t ≤ 3$$
The answer in my textbook says the minimum is $(3, -1)$ and the maximum is $(0, 2)$.
I keep getting different results, and I don't understand what I'm doing wrong. If someone could provide me with the general concept of how to solve these types of questions, I would really appreciate it.
My solution:
$f(-1) = 2-|-1|$
$f(-1) = 1$
$f(3) = 2 - |3|$
$f(3) = -1$
I thought doing that would give me the $y$ values for the absolute max and min, but clearly, it differs from the answer provided in the textbook.
Additionally, I don't really understand how to get the corresponding $x$ values for the max and min. I tried taking the derivative of the original function which left me with $f'(t) = -t/|t|$ but I didn't know where to go from there.
calculus derivatives absolute-value
calculus derivatives absolute-value
edited Nov 27 at 6:59
Eevee Trainer
2,466220
asked Nov 27 at 6:46
Blackthorn
174
edited Nov 27 at 6:59
Eevee Trainer
2,466220
asked Nov 27 at 6:46
Blackthorn
174
edited Nov 27 at 6:59
Eevee Trainer
2,466220
edited Nov 27 at 6:59
Eevee Trainer
2,466220
edited Nov 27 at 6:59
Eevee Trainer
2,466220
2,466220
asked Nov 27 at 6:46
Blackthorn
174
asked Nov 27 at 6:46
Blackthorn
174
asked Nov 27 at 6:46
Blackthorn
174
174
add a comment |
add a comment |
2 Answers
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0
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Hint: Split up $(-1,3)$ into two intervals: $(-1,0)$ and $(0,3)$. On the former interval, since $t<0$, $|t|=-t$ and thus $f(t) = 2 + t$ on that interval. Similarly, you can show $f(t) = 2 - t$ on the second interval.
Then consider the derivative of each function on each interval, and particularly what that derivative means in the scope of the function overall - derivative tests aren't going to help a whole lot here, so think moreso about what the derivative means in a qualitative sense and how it describes the behavior of the function.
Looking at a graph might prove useful for this qualitative analysis in particular so here's one I quickly hashed up on Desmos:
answered Nov 27 at 6:53
Eevee Trainer
2,466220
add a comment |
up vote
0
down vote
This is a constrained optimization problem with non-linear objective funciton. According to the Extreme Value Theorem, you must check the border points ($t=-1$, $t=3$) and the critical points in the constrained interval ($t=0$):
$$f(-1)=1;\
f(3)=-1 text{(min)};\
f(0)=2 text{(max)}.$$
answered Nov 27 at 7:27
farruhota
18.5k2736
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Hint: Split up $(-1,3)$ into two intervals: $(-1,0)$ and $(0,3)$. On the former interval, since $t<0$, $|t|=-t$ and thus $f(t) = 2 + t$ on that interval. Similarly, you can show $f(t) = 2 - t$ on the second interval.
Then consider the derivative of each function on each interval, and particularly what that derivative means in the scope of the function overall - derivative tests aren't going to help a whole lot here, so think moreso about what the derivative means in a qualitative sense and how it describes the behavior of the function.
Looking at a graph might prove useful for this qualitative analysis in particular so here's one I quickly hashed up on Desmos:
answered Nov 27 at 6:53
Eevee Trainer
2,466220
add a comment |
up vote
0
down vote
accepted
Hint: Split up $(-1,3)$ into two intervals: $(-1,0)$ and $(0,3)$. On the former interval, since $t<0$, $|t|=-t$ and thus $f(t) = 2 + t$ on that interval. Similarly, you can show $f(t) = 2 - t$ on the second interval.
Then consider the derivative of each function on each interval, and particularly what that derivative means in the scope of the function overall - derivative tests aren't going to help a whole lot here, so think moreso about what the derivative means in a qualitative sense and how it describes the behavior of the function.
Looking at a graph might prove useful for this qualitative analysis in particular so here's one I quickly hashed up on Desmos:
answered Nov 27 at 6:53
Eevee Trainer
2,466220
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Hint: Split up $(-1,3)$ into two intervals: $(-1,0)$ and $(0,3)$. On the former interval, since $t<0$, $|t|=-t$ and thus $f(t) = 2 + t$ on that interval. Similarly, you can show $f(t) = 2 - t$ on the second interval.
Then consider the derivative of each function on each interval, and particularly what that derivative means in the scope of the function overall - derivative tests aren't going to help a whole lot here, so think moreso about what the derivative means in a qualitative sense and how it describes the behavior of the function.
Looking at a graph might prove useful for this qualitative analysis in particular so here's one I quickly hashed up on Desmos:
answered Nov 27 at 6:53
Eevee Trainer
2,466220
Hint: Split up $(-1,3)$ into two intervals: $(-1,0)$ and $(0,3)$. On the former interval, since $t<0$, $|t|=-t$ and thus $f(t) = 2 + t$ on that interval. Similarly, you can show $f(t) = 2 - t$ on the second interval.
Then consider the derivative of each function on each interval, and particularly what that derivative means in the scope of the function overall - derivative tests aren't going to help a whole lot here, so think moreso about what the derivative means in a qualitative sense and how it describes the behavior of the function.
Looking at a graph might prove useful for this qualitative analysis in particular so here's one I quickly hashed up on Desmos:
answered Nov 27 at 6:53
Eevee Trainer
2,466220
answered Nov 27 at 6:53
Eevee Trainer
2,466220
answered Nov 27 at 6:53
Eevee Trainer
2,466220
answered Nov 27 at 6:53
Eevee Trainer
2,466220
2,466220
add a comment |
add a comment |
up vote
0
down vote
This is a constrained optimization problem with non-linear objective funciton. According to the Extreme Value Theorem, you must check the border points ($t=-1$, $t=3$) and the critical points in the constrained interval ($t=0$):
$$f(-1)=1;\
f(3)=-1 text{(min)};\
f(0)=2 text{(max)}.$$
answered Nov 27 at 7:27
farruhota
18.5k2736
add a comment |
up vote
0
down vote
This is a constrained optimization problem with non-linear objective funciton. According to the Extreme Value Theorem, you must check the border points ($t=-1$, $t=3$) and the critical points in the constrained interval ($t=0$):
$$f(-1)=1;\
f(3)=-1 text{(min)};\
f(0)=2 text{(max)}.$$
answered Nov 27 at 7:27
farruhota
18.5k2736
add a comment |
up vote
0
down vote
up vote
0
down vote
This is a constrained optimization problem with non-linear objective funciton. According to the Extreme Value Theorem, you must check the border points ($t=-1$, $t=3$) and the critical points in the constrained interval ($t=0$):
$$f(-1)=1;\
f(3)=-1 text{(min)};\
f(0)=2 text{(max)}.$$
answered Nov 27 at 7:27
farruhota
18.5k2736
This is a constrained optimization problem with non-linear objective funciton. According to the Extreme Value Theorem, you must check the border points ($t=-1$, $t=3$) and the critical points in the constrained interval ($t=0$):
$$f(-1)=1;\
f(3)=-1 text{(min)};\
f(0)=2 text{(max)}.$$
answered Nov 27 at 7:27
farruhota
18.5k2736
answered Nov 27 at 7:27
farruhota
18.5k2736
answered Nov 27 at 7:27
farruhota
18.5k2736
answered Nov 27 at 7:27
farruhota
18.5k2736
18.5k2736
add a comment |
add a comment |
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