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

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edited Aug 3 at 21:48

Math Lover

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asked Aug 3 at 21:19

bru1987

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up vote
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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

share|cite|improve this question

edited Aug 3 at 21:48

Math Lover

12.2k21132

asked Aug 3 at 21:19

bru1987

930818

add a comment | 

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

share|cite|improve this question

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

share|cite|improve this question

edited Aug 3 at 21:48

Math Lover

12.2k21132

asked Aug 3 at 21:19

bru1987

930818

share|cite|improve this question

share|cite|improve this question

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 | 

add a comment | 

1 Answer
1

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 $(-infty,8/3]$.

share|cite|improve this answer

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 | 

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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 $(-infty,8/3]$.

share|cite|improve this answer

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 | 

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 $(-infty,8/3]$.

share|cite|improve this answer

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 | 

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 $(-infty,8/3]$.

share|cite|improve this answer

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 $(-infty,8/3]$.

share|cite|improve this answer

answered Aug 3 at 21:23

egreg

164k1180187

share|cite|improve this answer

share|cite|improve this answer

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 | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 | 

 

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