Finding the minimum of an exponential function given its domain and maximum
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The domain of function $y=a^{vert(x-1)vert+2}$ is ${xvert-1le xle2}$, and the maximum value of the function (for the given domain) is $frac 14$. The problem is the find the minimum value of the function using the given information.
I looked at the solution for this problem and it went like this:
First they take the exponent of the original function as f(x):
$f(x)=vert(x-1)vert+2$
Using the given domain ${xvert-1le xle2}$, it can be deduced that $-2le x-1le1$. I understood everything until here, but then it goes on to say that from this it can also be deduced that
$0le vert x-1vertle2$
I really don't get how this can be deduced...
exponential-function
asked Nov 12 at 16:55
linnnn
665
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The domain of function $y=a^{vert(x-1)vert+2}$ is ${xvert-1le xle2}$, and the maximum value of the function (for the given domain) is $frac 14$. The problem is the find the minimum value of the function using the given information.
I looked at the solution for this problem and it went like this:
First they take the exponent of the original function as f(x):
$f(x)=vert(x-1)vert+2$
Using the given domain ${xvert-1le xle2}$, it can be deduced that $-2le x-1le1$. I understood everything until here, but then it goes on to say that from this it can also be deduced that
$0le vert x-1vertle2$
I really don't get how this can be deduced...
exponential-function
asked Nov 12 at 16:55
linnnn
665
Haha: A new problem for the CRUDE people and their disciples: Does regurgitating the full textbook solution count as enough "context" to save a question from the axe?
– Christian Blatter
Nov 12 at 19:20
add a comment |
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0
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up vote
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down vote
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The domain of function $y=a^{vert(x-1)vert+2}$ is ${xvert-1le xle2}$, and the maximum value of the function (for the given domain) is $frac 14$. The problem is the find the minimum value of the function using the given information.
I looked at the solution for this problem and it went like this:
First they take the exponent of the original function as f(x):
$f(x)=vert(x-1)vert+2$
Using the given domain ${xvert-1le xle2}$, it can be deduced that $-2le x-1le1$. I understood everything until here, but then it goes on to say that from this it can also be deduced that
$0le vert x-1vertle2$
I really don't get how this can be deduced...
exponential-function
asked Nov 12 at 16:55
linnnn
665
The domain of function $y=a^{vert(x-1)vert+2}$ is ${xvert-1le xle2}$, and the maximum value of the function (for the given domain) is $frac 14$. The problem is the find the minimum value of the function using the given information.
I looked at the solution for this problem and it went like this:
First they take the exponent of the original function as f(x):
$f(x)=vert(x-1)vert+2$
Using the given domain ${xvert-1le xle2}$, it can be deduced that $-2le x-1le1$. I understood everything until here, but then it goes on to say that from this it can also be deduced that
$0le vert x-1vertle2$
I really don't get how this can be deduced...
exponential-function
exponential-function
asked Nov 12 at 16:55
linnnn
665
asked Nov 12 at 16:55
linnnn
665
asked Nov 12 at 16:55
linnnn
665
asked Nov 12 at 16:55
linnnn
665
asked Nov 12 at 16:55
linnnn
665
665
Haha: A new problem for the CRUDE people and their disciples: Does regurgitating the full textbook solution count as enough "context" to save a question from the axe?
– Christian Blatter
Nov 12 at 19:20
add a comment |
Haha: A new problem for the CRUDE people and their disciples: Does regurgitating the full textbook solution count as enough "context" to save a question from the axe?
– Christian Blatter
Nov 12 at 19:20
Haha: A new problem for the CRUDE people and their disciples: Does regurgitating the full textbook solution count as enough "context" to save a question from the axe?
– Christian Blatter
Nov 12 at 19:20
Haha: A new problem for the CRUDE people and their disciples: Does regurgitating the full textbook solution count as enough "context" to save a question from the axe?
– Christian Blatter
Nov 12 at 19:20
add a comment |
1 Answer
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Draw $y mapsto |y|$. What's the image of $[-2; 1]$?
answered Nov 12 at 16:58
Stockfish
40426
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Draw $y mapsto |y|$. What's the image of $[-2; 1]$?
answered Nov 12 at 16:58
Stockfish
40426
add a comment |
up vote
0
down vote
Draw $y mapsto |y|$. What's the image of $[-2; 1]$?
answered Nov 12 at 16:58
Stockfish
40426
add a comment |
up vote
0
down vote
up vote
0
down vote
Draw $y mapsto |y|$. What's the image of $[-2; 1]$?
answered Nov 12 at 16:58
Stockfish
40426
Draw $y mapsto |y|$. What's the image of $[-2; 1]$?
answered Nov 12 at 16:58
Stockfish
40426
answered Nov 12 at 16:58
Stockfish
40426
answered Nov 12 at 16:58
Stockfish
40426
answered Nov 12 at 16:58
Stockfish
40426
40426
add a comment |
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Haha: A new problem for the CRUDE people and their disciples: Does regurgitating the full textbook solution count as enough "context" to save a question from the axe?
– Christian Blatter
Nov 12 at 19:20