Mixing
Find $f([-1,2])$ if $f:Rto R$ and $f(x)=x^2$.
Clash Royale CLAN TAG#URR8PPP
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May someone confirm if im right on this exercise:
I gotta find $f([-1,2])$ if $f:Rto R$ and $f(x)=x^2$
So is the answer $[1,4]$?
calculus functions
edited Aug 2 at 11:41
user 108128
18.8k41544
asked Aug 2 at 11:39
MZ97
237
add a comment |Â
up vote
0
down vote
favorite
May someone confirm if im right on this exercise:
I gotta find $f([-1,2])$ if $f:Rto R$ and $f(x)=x^2$
So is the answer $[1,4]$?
calculus functions
edited Aug 2 at 11:41
user 108128
18.8k41544
asked Aug 2 at 11:39
MZ97
237
incorrect......
– user 108128
Aug 2 at 11:40
2
I guess here you're supposed to find the minimum and maximum of the function in the given interval. So you don't just stick the end values to the function and hope for the best.
– Matti P.
Aug 2 at 11:42
4
Observe that $f(0)=0$
– Gustave
Aug 2 at 11:42
hmm, then i need more help
– MZ97
Aug 2 at 11:43
Are you familiar with monotonic functions ?
– nicomezi
Aug 2 at 11:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
May someone confirm if im right on this exercise:
I gotta find $f([-1,2])$ if $f:Rto R$ and $f(x)=x^2$
So is the answer $[1,4]$?
calculus functions
edited Aug 2 at 11:41
user 108128
18.8k41544
asked Aug 2 at 11:39
MZ97
237
May someone confirm if im right on this exercise:
I gotta find $f([-1,2])$ if $f:Rto R$ and $f(x)=x^2$
So is the answer $[1,4]$?
calculus functions
edited Aug 2 at 11:41
user 108128
18.8k41544
asked Aug 2 at 11:39
MZ97
237
edited Aug 2 at 11:41
user 108128
18.8k41544
edited Aug 2 at 11:41
user 108128
18.8k41544
edited Aug 2 at 11:41
user 108128
18.8k41544
18.8k41544
asked Aug 2 at 11:39
MZ97
237
asked Aug 2 at 11:39
MZ97
237
asked Aug 2 at 11:39
MZ97
237
237
incorrect......
– user 108128
Aug 2 at 11:40
2
I guess here you're supposed to find the minimum and maximum of the function in the given interval. So you don't just stick the end values to the function and hope for the best.
– Matti P.
Aug 2 at 11:42
4
Observe that $f(0)=0$
– Gustave
Aug 2 at 11:42
hmm, then i need more help
– MZ97
Aug 2 at 11:43
Are you familiar with monotonic functions ?
– nicomezi
Aug 2 at 11:47
add a comment |Â
incorrect......
– user 108128
Aug 2 at 11:40
2
I guess here you're supposed to find the minimum and maximum of the function in the given interval. So you don't just stick the end values to the function and hope for the best.
– Matti P.
Aug 2 at 11:42
4
Observe that $f(0)=0$
– Gustave
Aug 2 at 11:42
hmm, then i need more help
– MZ97
Aug 2 at 11:43
Are you familiar with monotonic functions ?
– nicomezi
Aug 2 at 11:47
incorrect......
– user 108128
Aug 2 at 11:40
incorrect......
– user 108128
Aug 2 at 11:40
2
2
I guess here you're supposed to find the minimum and maximum of the function in the given interval. So you don't just stick the end values to the function and hope for the best.
– Matti P.
Aug 2 at 11:42
I guess here you're supposed to find the minimum and maximum of the function in the given interval. So you don't just stick the end values to the function and hope for the best.
– Matti P.
Aug 2 at 11:42
4
4
Observe that $f(0)=0$
– Gustave
Aug 2 at 11:42
Observe that $f(0)=0$
– Gustave
Aug 2 at 11:42
hmm, then i need more help
– MZ97
Aug 2 at 11:43
hmm, then i need more help
– MZ97
Aug 2 at 11:43
Are you familiar with monotonic functions ?
– nicomezi
Aug 2 at 11:47
Are you familiar with monotonic functions ?
– nicomezi
Aug 2 at 11:47
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
See this graph so you can get were you problem is.
Finding $f([-1,2])$ means finding all the values $f(x)$ takes when $x$ takes values in $[-1,2]$
What you did was calculating $f(-1)$ and $f(2)$. It would have been correct if $f$ was monotonic on $[-1,2]$ (always increasing, or always decreasing).
However, as you can see from the graph (and maybe already knew), $x rightarrow x²$ is decreasing on$]-infty,0]$ and increasing on $[0, +infty[$.
So as for user108128 answer, you have to work on $[-1,0]$ and $[0,2]$. $f$ being monotonic on these intervals, you can then use your method.
answered Aug 2 at 12:09
F.Carette
1717
add a comment |Â
up vote
1
down vote
More help is with $-1leq xleq2$ then you have two intervals $[-1,0]cup[0,2]$, now for these intervals $f([-1,0])=[0,1]$ and $f([0,2])=[0,4]$. Then $f([-1,2])=[0,1]cup[0,4]=[0,4]$.
answered Aug 2 at 11:48
user 108128
18.8k41544
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
See this graph so you can get were you problem is.
Finding $f([-1,2])$ means finding all the values $f(x)$ takes when $x$ takes values in $[-1,2]$
What you did was calculating $f(-1)$ and $f(2)$. It would have been correct if $f$ was monotonic on $[-1,2]$ (always increasing, or always decreasing).
However, as you can see from the graph (and maybe already knew), $x rightarrow x²$ is decreasing on$]-infty,0]$ and increasing on $[0, +infty[$.
So as for user108128 answer, you have to work on $[-1,0]$ and $[0,2]$. $f$ being monotonic on these intervals, you can then use your method.
answered Aug 2 at 12:09
F.Carette
1717
add a comment |Â
up vote
3
down vote
See this graph so you can get were you problem is.
Finding $f([-1,2])$ means finding all the values $f(x)$ takes when $x$ takes values in $[-1,2]$
What you did was calculating $f(-1)$ and $f(2)$. It would have been correct if $f$ was monotonic on $[-1,2]$ (always increasing, or always decreasing).
However, as you can see from the graph (and maybe already knew), $x rightarrow x²$ is decreasing on$]-infty,0]$ and increasing on $[0, +infty[$.
So as for user108128 answer, you have to work on $[-1,0]$ and $[0,2]$. $f$ being monotonic on these intervals, you can then use your method.
answered Aug 2 at 12:09
F.Carette
1717
add a comment |Â
up vote
3
down vote
up vote
3
down vote
See this graph so you can get were you problem is.
Finding $f([-1,2])$ means finding all the values $f(x)$ takes when $x$ takes values in $[-1,2]$
What you did was calculating $f(-1)$ and $f(2)$. It would have been correct if $f$ was monotonic on $[-1,2]$ (always increasing, or always decreasing).
However, as you can see from the graph (and maybe already knew), $x rightarrow x²$ is decreasing on$]-infty,0]$ and increasing on $[0, +infty[$.
So as for user108128 answer, you have to work on $[-1,0]$ and $[0,2]$. $f$ being monotonic on these intervals, you can then use your method.
answered Aug 2 at 12:09
F.Carette
1717
See this graph so you can get were you problem is.
Finding $f([-1,2])$ means finding all the values $f(x)$ takes when $x$ takes values in $[-1,2]$
What you did was calculating $f(-1)$ and $f(2)$. It would have been correct if $f$ was monotonic on $[-1,2]$ (always increasing, or always decreasing).
However, as you can see from the graph (and maybe already knew), $x rightarrow x²$ is decreasing on$]-infty,0]$ and increasing on $[0, +infty[$.
So as for user108128 answer, you have to work on $[-1,0]$ and $[0,2]$. $f$ being monotonic on these intervals, you can then use your method.
answered Aug 2 at 12:09
F.Carette
1717
answered Aug 2 at 12:09
F.Carette
1717
answered Aug 2 at 12:09
F.Carette
1717
answered Aug 2 at 12:09
F.Carette
1717
1717
add a comment |Â
add a comment |Â
up vote
1
down vote
More help is with $-1leq xleq2$ then you have two intervals $[-1,0]cup[0,2]$, now for these intervals $f([-1,0])=[0,1]$ and $f([0,2])=[0,4]$. Then $f([-1,2])=[0,1]cup[0,4]=[0,4]$.
answered Aug 2 at 11:48
user 108128
18.8k41544
add a comment |Â
up vote
1
down vote
More help is with $-1leq xleq2$ then you have two intervals $[-1,0]cup[0,2]$, now for these intervals $f([-1,0])=[0,1]$ and $f([0,2])=[0,4]$. Then $f([-1,2])=[0,1]cup[0,4]=[0,4]$.
answered Aug 2 at 11:48
user 108128
18.8k41544
add a comment |Â
up vote
1
down vote
up vote
1
down vote
More help is with $-1leq xleq2$ then you have two intervals $[-1,0]cup[0,2]$, now for these intervals $f([-1,0])=[0,1]$ and $f([0,2])=[0,4]$. Then $f([-1,2])=[0,1]cup[0,4]=[0,4]$.
answered Aug 2 at 11:48
user 108128
18.8k41544
More help is with $-1leq xleq2$ then you have two intervals $[-1,0]cup[0,2]$, now for these intervals $f([-1,0])=[0,1]$ and $f([0,2])=[0,4]$. Then $f([-1,2])=[0,1]cup[0,4]=[0,4]$.
answered Aug 2 at 11:48
user 108128
18.8k41544
answered Aug 2 at 11:48
user 108128
18.8k41544
answered Aug 2 at 11:48
user 108128
18.8k41544
answered Aug 2 at 11:48
user 108128
18.8k41544
18.8k41544
add a comment |Â
add a comment |Â
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incorrect......
– user 108128
Aug 2 at 11:40
2
I guess here you're supposed to find the minimum and maximum of the function in the given interval. So you don't just stick the end values to the function and hope for the best.
– Matti P.
Aug 2 at 11:42
4
Observe that $f(0)=0$
– Gustave
Aug 2 at 11:42
hmm, then i need more help
– MZ97
Aug 2 at 11:43
Are you familiar with monotonic functions ?
– nicomezi
Aug 2 at 11:47