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Is $f(x) = x sqrt4-x$ decreasing at $x = 4$?
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I am solving a single variable calculus problem and it asks me to determine the decreasing and decreasing intervals of the function $f(x) = x sqrt4-x$. It's pretty clear that from $left] -infty , frac83 right[ $ the function is increasing but, since the domain of the function goes only until $x = 4$, should I write that the decreasing interval of the function is $left] frac83 , 4 right[$ or $left] frac83 , 4 right]$?
calculus maxima-minima
edited Aug 3 at 21:48
Math Lover
12.2k21132
asked Aug 3 at 21:19
bru1987
930818
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up vote
1
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I am solving a single variable calculus problem and it asks me to determine the decreasing and decreasing intervals of the function $f(x) = x sqrt4-x$. It's pretty clear that from $left] -infty , frac83 right[ $ the function is increasing but, since the domain of the function goes only until $x = 4$, should I write that the decreasing interval of the function is $left] frac83 , 4 right[$ or $left] frac83 , 4 right]$?
calculus maxima-minima
edited Aug 3 at 21:48
Math Lover
12.2k21132
asked Aug 3 at 21:19
bru1987
930818
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am solving a single variable calculus problem and it asks me to determine the decreasing and decreasing intervals of the function $f(x) = x sqrt4-x$. It's pretty clear that from $left] -infty , frac83 right[ $ the function is increasing but, since the domain of the function goes only until $x = 4$, should I write that the decreasing interval of the function is $left] frac83 , 4 right[$ or $left] frac83 , 4 right]$?
calculus maxima-minima
edited Aug 3 at 21:48
Math Lover
12.2k21132
asked Aug 3 at 21:19
bru1987
930818
I am solving a single variable calculus problem and it asks me to determine the decreasing and decreasing intervals of the function $f(x) = x sqrt4-x$. It's pretty clear that from $left] -infty , frac83 right[ $ the function is increasing but, since the domain of the function goes only until $x = 4$, should I write that the decreasing interval of the function is $left] frac83 , 4 right[$ or $left] frac83 , 4 right]$?
calculus maxima-minima
edited Aug 3 at 21:48
Math Lover
12.2k21132
asked Aug 3 at 21:19
bru1987
930818
edited Aug 3 at 21:48
Math Lover
12.2k21132
edited Aug 3 at 21:48
Math Lover
12.2k21132
edited Aug 3 at 21:48
Math Lover
12.2k21132
12.2k21132
asked Aug 3 at 21:19
bru1987
930818
asked Aug 3 at 21:19
bru1987
930818
asked Aug 3 at 21:19
bru1987
930818
930818
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
Neither.
The derivative of your function $f$ is
$$
f'(x)=sqrt4-x-fracx2sqrt4-x=frac8-3x2sqrt4-x
$$
for $x<4$ (no differentiability at $4$).
The point $x=8/3$ is a local maximum. Hence the function is decreasing over
$$
[8/3,4]
$$
because, for every $x,yin [8/3,4]$, if $x<y$ then $f(x)>f(y)$.
The function is increasing over $(-infty,8/3]$.
answered Aug 3 at 21:23
egreg
164k1180187
thank you for your input. So I include 8/3 on both intervals? That seems weird - shouldn't I disregard it on both intervals since precisely at 8/3 the function is neither increasing nor decreasing?
– bru1987
Aug 3 at 21:29
1
@bru1987 At a single point the function is neither increasing nor decreasing: this notion makes sense over intervals. There is no problem if a point belongs to both the intervals where the function is increasing and decreasing.
– egreg
Aug 3 at 21:31
I understand your explanation, but it still seems off. Let me put it another way: would it be actually wrong to use the intervals $]-infty, 8/3[$ and $]8/3,4]$?
– bru1987
Aug 3 at 21:33
1
@bru1987 Yes: the function is decreasing over $[8/3,4]$ and you're missing one point.
– egreg
Aug 3 at 21:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Neither.
The derivative of your function $f$ is
$$
f'(x)=sqrt4-x-fracx2sqrt4-x=frac8-3x2sqrt4-x
$$
for $x<4$ (no differentiability at $4$).
The point $x=8/3$ is a local maximum. Hence the function is decreasing over
$$
[8/3,4]
$$
because, for every $x,yin [8/3,4]$, if $x<y$ then $f(x)>f(y)$.
The function is increasing over $(-infty,8/3]$.
answered Aug 3 at 21:23
egreg
164k1180187
thank you for your input. So I include 8/3 on both intervals? That seems weird - shouldn't I disregard it on both intervals since precisely at 8/3 the function is neither increasing nor decreasing?
– bru1987
Aug 3 at 21:29
1
@bru1987 At a single point the function is neither increasing nor decreasing: this notion makes sense over intervals. There is no problem if a point belongs to both the intervals where the function is increasing and decreasing.
– egreg
Aug 3 at 21:31
I understand your explanation, but it still seems off. Let me put it another way: would it be actually wrong to use the intervals $]-infty, 8/3[$ and $]8/3,4]$?
– bru1987
Aug 3 at 21:33
1
@bru1987 Yes: the function is decreasing over $[8/3,4]$ and you're missing one point.
– egreg
Aug 3 at 21:37
add a comment |Â
up vote
2
down vote
accepted
Neither.
The derivative of your function $f$ is
$$
f'(x)=sqrt4-x-fracx2sqrt4-x=frac8-3x2sqrt4-x
$$
for $x<4$ (no differentiability at $4$).
The point $x=8/3$ is a local maximum. Hence the function is decreasing over
$$
[8/3,4]
$$
because, for every $x,yin [8/3,4]$, if $x<y$ then $f(x)>f(y)$.
The function is increasing over $(-infty,8/3]$.
answered Aug 3 at 21:23
egreg
164k1180187
thank you for your input. So I include 8/3 on both intervals? That seems weird - shouldn't I disregard it on both intervals since precisely at 8/3 the function is neither increasing nor decreasing?
– bru1987
Aug 3 at 21:29
1
@bru1987 At a single point the function is neither increasing nor decreasing: this notion makes sense over intervals. There is no problem if a point belongs to both the intervals where the function is increasing and decreasing.
– egreg
Aug 3 at 21:31
I understand your explanation, but it still seems off. Let me put it another way: would it be actually wrong to use the intervals $]-infty, 8/3[$ and $]8/3,4]$?
– bru1987
Aug 3 at 21:33
1
@bru1987 Yes: the function is decreasing over $[8/3,4]$ and you're missing one point.
– egreg
Aug 3 at 21:37
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Neither.
The derivative of your function $f$ is
$$
f'(x)=sqrt4-x-fracx2sqrt4-x=frac8-3x2sqrt4-x
$$
for $x<4$ (no differentiability at $4$).
The point $x=8/3$ is a local maximum. Hence the function is decreasing over
$$
[8/3,4]
$$
because, for every $x,yin [8/3,4]$, if $x<y$ then $f(x)>f(y)$.
The function is increasing over $(-infty,8/3]$.
answered Aug 3 at 21:23
egreg
164k1180187
Neither.
The derivative of your function $f$ is
$$
f'(x)=sqrt4-x-fracx2sqrt4-x=frac8-3x2sqrt4-x
$$
for $x<4$ (no differentiability at $4$).
The point $x=8/3$ is a local maximum. Hence the function is decreasing over
$$
[8/3,4]
$$
because, for every $x,yin [8/3,4]$, if $x<y$ then $f(x)>f(y)$.
The function is increasing over $(-infty,8/3]$.
answered Aug 3 at 21:23
egreg
164k1180187
answered Aug 3 at 21:23
egreg
164k1180187
answered Aug 3 at 21:23
egreg
164k1180187
answered Aug 3 at 21:23
egreg
164k1180187
164k1180187
thank you for your input. So I include 8/3 on both intervals? That seems weird - shouldn't I disregard it on both intervals since precisely at 8/3 the function is neither increasing nor decreasing?
– bru1987
Aug 3 at 21:29
1
@bru1987 At a single point the function is neither increasing nor decreasing: this notion makes sense over intervals. There is no problem if a point belongs to both the intervals where the function is increasing and decreasing.
– egreg
Aug 3 at 21:31
I understand your explanation, but it still seems off. Let me put it another way: would it be actually wrong to use the intervals $]-infty, 8/3[$ and $]8/3,4]$?
– bru1987
Aug 3 at 21:33
1
@bru1987 Yes: the function is decreasing over $[8/3,4]$ and you're missing one point.
– egreg
Aug 3 at 21:37
add a comment |Â
thank you for your input. So I include 8/3 on both intervals? That seems weird - shouldn't I disregard it on both intervals since precisely at 8/3 the function is neither increasing nor decreasing?
– bru1987
Aug 3 at 21:29
1
@bru1987 At a single point the function is neither increasing nor decreasing: this notion makes sense over intervals. There is no problem if a point belongs to both the intervals where the function is increasing and decreasing.
– egreg
Aug 3 at 21:31
I understand your explanation, but it still seems off. Let me put it another way: would it be actually wrong to use the intervals $]-infty, 8/3[$ and $]8/3,4]$?
– bru1987
Aug 3 at 21:33
1
@bru1987 Yes: the function is decreasing over $[8/3,4]$ and you're missing one point.
– egreg
Aug 3 at 21:37
thank you for your input. So I include 8/3 on both intervals? That seems weird - shouldn't I disregard it on both intervals since precisely at 8/3 the function is neither increasing nor decreasing?
– bru1987
Aug 3 at 21:29
thank you for your input. So I include 8/3 on both intervals? That seems weird - shouldn't I disregard it on both intervals since precisely at 8/3 the function is neither increasing nor decreasing?
– bru1987
Aug 3 at 21:29
1
1
@bru1987 At a single point the function is neither increasing nor decreasing: this notion makes sense over intervals. There is no problem if a point belongs to both the intervals where the function is increasing and decreasing.
– egreg
Aug 3 at 21:31
@bru1987 At a single point the function is neither increasing nor decreasing: this notion makes sense over intervals. There is no problem if a point belongs to both the intervals where the function is increasing and decreasing.
– egreg
Aug 3 at 21:31
I understand your explanation, but it still seems off. Let me put it another way: would it be actually wrong to use the intervals $]-infty, 8/3[$ and $]8/3,4]$?
– bru1987
Aug 3 at 21:33
I understand your explanation, but it still seems off. Let me put it another way: would it be actually wrong to use the intervals $]-infty, 8/3[$ and $]8/3,4]$?
– bru1987
Aug 3 at 21:33
1
1
@bru1987 Yes: the function is decreasing over $[8/3,4]$ and you're missing one point.
– egreg
Aug 3 at 21:37
@bru1987 Yes: the function is decreasing over $[8/3,4]$ and you're missing one point.
– egreg
Aug 3 at 21:37
add a comment |Â
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