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Integration of a function provided as an implicit function [closed]
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Let a differentiable function $f$ satisfies the functional rule $$f(xy) = f(x) + f(y) + xy - x -y $$ for all values of $x,y > 0$ and $f'(1)=4$.
Based on this statement there were three questions:
If $f(x_0)=0$, then $x_0$ lies in the interval
$$
(a) (0,1) \
(b) (1,e) \
(c) (e, e^2) \
(d) (e^2,e^3)
$$$intfracf(x)xdx$ is equal to:
$$
(a) 3(ln x)^2 + x + c \
(b) 3(ln x) + x + c \
(c) 1.5(ln x)^2 + x + c \
(d) 1.5(ln x) + x + c
$$If $inte^f(x)dx = e^x(ax^3 + bx^2 + cx + d) + lambda$, then the value of $(a+b+c+d)$ is equal to:
$$
(a) -1 \
(b) -2 \
(c) 3 \
(d) 6
$$
My attempt : I had no idea how to deal with implicit function. Any hints or suggestion to handle this kind of question would have been extremely helpful. Thanks for giving me the pointers. I will try and update the full solutions soon.
integration indefinite-integrals implicit-function
asked Jul 20 at 5:47
MathsLearner
657213
closed as off-topic by heropup, Trần Thúc Minh TrÃ, José Carlos Santos, amWhy, Parcly Taxel Jul 20 at 14:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Trần Thúc Minh TrÃÂ, José Carlos Santos, amWhy, Parcly Taxel
add a comment |Â
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Let a differentiable function $f$ satisfies the functional rule $$f(xy) = f(x) + f(y) + xy - x -y $$ for all values of $x,y > 0$ and $f'(1)=4$.
Based on this statement there were three questions:
If $f(x_0)=0$, then $x_0$ lies in the interval
$$
(a) (0,1) \
(b) (1,e) \
(c) (e, e^2) \
(d) (e^2,e^3)
$$$intfracf(x)xdx$ is equal to:
$$
(a) 3(ln x)^2 + x + c \
(b) 3(ln x) + x + c \
(c) 1.5(ln x)^2 + x + c \
(d) 1.5(ln x) + x + c
$$If $inte^f(x)dx = e^x(ax^3 + bx^2 + cx + d) + lambda$, then the value of $(a+b+c+d)$ is equal to:
$$
(a) -1 \
(b) -2 \
(c) 3 \
(d) 6
$$
My attempt : I had no idea how to deal with implicit function. Any hints or suggestion to handle this kind of question would have been extremely helpful. Thanks for giving me the pointers. I will try and update the full solutions soon.
integration indefinite-integrals implicit-function
asked Jul 20 at 5:47
MathsLearner
657213
closed as off-topic by heropup, Trần Thúc Minh TrÃ, José Carlos Santos, amWhy, Parcly Taxel Jul 20 at 14:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Trần Thúc Minh TrÃÂ, José Carlos Santos, amWhy, Parcly Taxel
1
It's possible to solve the equation for $f$. Differentiate the equation wrt to $x$ holding $y$ fixed.
– Oliver Jones
Jul 20 at 6:20
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let a differentiable function $f$ satisfies the functional rule $$f(xy) = f(x) + f(y) + xy - x -y $$ for all values of $x,y > 0$ and $f'(1)=4$.
Based on this statement there were three questions:
If $f(x_0)=0$, then $x_0$ lies in the interval
$$
(a) (0,1) \
(b) (1,e) \
(c) (e, e^2) \
(d) (e^2,e^3)
$$$intfracf(x)xdx$ is equal to:
$$
(a) 3(ln x)^2 + x + c \
(b) 3(ln x) + x + c \
(c) 1.5(ln x)^2 + x + c \
(d) 1.5(ln x) + x + c
$$If $inte^f(x)dx = e^x(ax^3 + bx^2 + cx + d) + lambda$, then the value of $(a+b+c+d)$ is equal to:
$$
(a) -1 \
(b) -2 \
(c) 3 \
(d) 6
$$
My attempt : I had no idea how to deal with implicit function. Any hints or suggestion to handle this kind of question would have been extremely helpful. Thanks for giving me the pointers. I will try and update the full solutions soon.
integration indefinite-integrals implicit-function
asked Jul 20 at 5:47
MathsLearner
657213
Let a differentiable function $f$ satisfies the functional rule $$f(xy) = f(x) + f(y) + xy - x -y $$ for all values of $x,y > 0$ and $f'(1)=4$.
Based on this statement there were three questions:
If $f(x_0)=0$, then $x_0$ lies in the interval
$$
(a) (0,1) \
(b) (1,e) \
(c) (e, e^2) \
(d) (e^2,e^3)
$$$intfracf(x)xdx$ is equal to:
$$
(a) 3(ln x)^2 + x + c \
(b) 3(ln x) + x + c \
(c) 1.5(ln x)^2 + x + c \
(d) 1.5(ln x) + x + c
$$If $inte^f(x)dx = e^x(ax^3 + bx^2 + cx + d) + lambda$, then the value of $(a+b+c+d)$ is equal to:
$$
(a) -1 \
(b) -2 \
(c) 3 \
(d) 6
$$
My attempt : I had no idea how to deal with implicit function. Any hints or suggestion to handle this kind of question would have been extremely helpful. Thanks for giving me the pointers. I will try and update the full solutions soon.
integration indefinite-integrals implicit-function
asked Jul 20 at 5:47
MathsLearner
657213
edited Jul 24 at 13:35
asked Jul 20 at 5:47
MathsLearner
657213
asked Jul 20 at 5:47
MathsLearner
657213
asked Jul 20 at 5:47
MathsLearner
657213
657213
closed as off-topic by heropup, Trần Thúc Minh TrÃ, José Carlos Santos, amWhy, Parcly Taxel Jul 20 at 14:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Trần Thúc Minh TrÃÂ, José Carlos Santos, amWhy, Parcly Taxel
closed as off-topic by heropup, Trần Thúc Minh TrÃ, José Carlos Santos, amWhy, Parcly Taxel Jul 20 at 14:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Trần Thúc Minh TrÃÂ, José Carlos Santos, amWhy, Parcly Taxel
1
It's possible to solve the equation for $f$. Differentiate the equation wrt to $x$ holding $y$ fixed.
– Oliver Jones
Jul 20 at 6:20
add a comment |Â
1
It's possible to solve the equation for $f$. Differentiate the equation wrt to $x$ holding $y$ fixed.
– Oliver Jones
Jul 20 at 6:20
1
1
It's possible to solve the equation for $f$. Differentiate the equation wrt to $x$ holding $y$ fixed.
– Oliver Jones
Jul 20 at 6:20
It's possible to solve the equation for $f$. Differentiate the equation wrt to $x$ holding $y$ fixed.
– Oliver Jones
Jul 20 at 6:20
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
Differentiate $f(xy) = f(x) + f(y) + xy - x -y$ of the $y$ variable:
$$
xf'(xy)=f'(y)+x-1.
$$
Let $y=1$ then we get
$$
xf'(x)=f'(1)+x-1
$$
Since $f'(1)=4$ we get
$$
f'(x)=frac3x+1
$$
Hence
$$
f(x)=3ln x +x+c
$$
where $c$ is an arbitrary constant.
Let $x=y=1$ in the original equation we get $f(1)=1$, hence we can get $c=0$, hence
$$
f(x)=3ln x+x.
$$
Take this into the equation to examine it satisfies the condition.
Now you can solve the question directly.
answered Jul 20 at 6:23
yahoo
536312
1
You have to check that the $f$ you found actually solves the given functional equation.
– Christian Blatter
Jul 20 at 8:16
@Chr Yes, it is necessary to check $f$ satisfy the condition, thank you.
– yahoo
Jul 20 at 8:33
add a comment |Â
up vote
0
down vote
The equation itself may be solved without any regularity conditions by observing that $g$, $g(x):=f(x)-x$ satisfies the logarithmic equation $g(xy)=g(x)+g(y)$ for $gcolon (0,infty)tomathbbR$. Thus there is some additive function $acolonmathbbRtomathbbR$, i.e., $a$ satisfies $a(x+y)=a(x)+a(y)$ for all real $x,y$, such that $g(x)=a(ln x)$ for all $x>0$. Thus $f$ with $f(x):=a(ln x)+x$ is the general solutions. If $f$ is assumed to be regular, for example continuous, the function $a$ has to be of the form $xmapsto c x$ with some constant $c$. Thus in particular holde true If $f$ is differentiable. In that case the condition on $f'$ results in the form of $f$ as given in the other answer.
answered Jul 20 at 10:54
Jens Schwaiger
1,112116
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Differentiate $f(xy) = f(x) + f(y) + xy - x -y$ of the $y$ variable:
$$
xf'(xy)=f'(y)+x-1.
$$
Let $y=1$ then we get
$$
xf'(x)=f'(1)+x-1
$$
Since $f'(1)=4$ we get
$$
f'(x)=frac3x+1
$$
Hence
$$
f(x)=3ln x +x+c
$$
where $c$ is an arbitrary constant.
Let $x=y=1$ in the original equation we get $f(1)=1$, hence we can get $c=0$, hence
$$
f(x)=3ln x+x.
$$
Take this into the equation to examine it satisfies the condition.
Now you can solve the question directly.
answered Jul 20 at 6:23
yahoo
536312
1
You have to check that the $f$ you found actually solves the given functional equation.
– Christian Blatter
Jul 20 at 8:16
@Chr Yes, it is necessary to check $f$ satisfy the condition, thank you.
– yahoo
Jul 20 at 8:33
add a comment |Â
up vote
1
down vote
Differentiate $f(xy) = f(x) + f(y) + xy - x -y$ of the $y$ variable:
$$
xf'(xy)=f'(y)+x-1.
$$
Let $y=1$ then we get
$$
xf'(x)=f'(1)+x-1
$$
Since $f'(1)=4$ we get
$$
f'(x)=frac3x+1
$$
Hence
$$
f(x)=3ln x +x+c
$$
where $c$ is an arbitrary constant.
Let $x=y=1$ in the original equation we get $f(1)=1$, hence we can get $c=0$, hence
$$
f(x)=3ln x+x.
$$
Take this into the equation to examine it satisfies the condition.
Now you can solve the question directly.
answered Jul 20 at 6:23
yahoo
536312
1
You have to check that the $f$ you found actually solves the given functional equation.
– Christian Blatter
Jul 20 at 8:16
@Chr Yes, it is necessary to check $f$ satisfy the condition, thank you.
– yahoo
Jul 20 at 8:33
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Differentiate $f(xy) = f(x) + f(y) + xy - x -y$ of the $y$ variable:
$$
xf'(xy)=f'(y)+x-1.
$$
Let $y=1$ then we get
$$
xf'(x)=f'(1)+x-1
$$
Since $f'(1)=4$ we get
$$
f'(x)=frac3x+1
$$
Hence
$$
f(x)=3ln x +x+c
$$
where $c$ is an arbitrary constant.
Let $x=y=1$ in the original equation we get $f(1)=1$, hence we can get $c=0$, hence
$$
f(x)=3ln x+x.
$$
Take this into the equation to examine it satisfies the condition.
Now you can solve the question directly.
answered Jul 20 at 6:23
yahoo
536312
Differentiate $f(xy) = f(x) + f(y) + xy - x -y$ of the $y$ variable:
$$
xf'(xy)=f'(y)+x-1.
$$
Let $y=1$ then we get
$$
xf'(x)=f'(1)+x-1
$$
Since $f'(1)=4$ we get
$$
f'(x)=frac3x+1
$$
Hence
$$
f(x)=3ln x +x+c
$$
where $c$ is an arbitrary constant.
Let $x=y=1$ in the original equation we get $f(1)=1$, hence we can get $c=0$, hence
$$
f(x)=3ln x+x.
$$
Take this into the equation to examine it satisfies the condition.
Now you can solve the question directly.
answered Jul 20 at 6:23
yahoo
536312
edited Jul 20 at 9:00
answered Jul 20 at 6:23
yahoo
536312
answered Jul 20 at 6:23
yahoo
536312
answered Jul 20 at 6:23
yahoo
536312
536312
1
You have to check that the $f$ you found actually solves the given functional equation.
– Christian Blatter
Jul 20 at 8:16
@Chr Yes, it is necessary to check $f$ satisfy the condition, thank you.
– yahoo
Jul 20 at 8:33
add a comment |Â
1
You have to check that the $f$ you found actually solves the given functional equation.
– Christian Blatter
Jul 20 at 8:16
@Chr Yes, it is necessary to check $f$ satisfy the condition, thank you.
– yahoo
Jul 20 at 8:33
1
1
You have to check that the $f$ you found actually solves the given functional equation.
– Christian Blatter
Jul 20 at 8:16
You have to check that the $f$ you found actually solves the given functional equation.
– Christian Blatter
Jul 20 at 8:16
@Chr Yes, it is necessary to check $f$ satisfy the condition, thank you.
– yahoo
Jul 20 at 8:33
@Chr Yes, it is necessary to check $f$ satisfy the condition, thank you.
– yahoo
Jul 20 at 8:33
add a comment |Â
up vote
0
down vote
The equation itself may be solved without any regularity conditions by observing that $g$, $g(x):=f(x)-x$ satisfies the logarithmic equation $g(xy)=g(x)+g(y)$ for $gcolon (0,infty)tomathbbR$. Thus there is some additive function $acolonmathbbRtomathbbR$, i.e., $a$ satisfies $a(x+y)=a(x)+a(y)$ for all real $x,y$, such that $g(x)=a(ln x)$ for all $x>0$. Thus $f$ with $f(x):=a(ln x)+x$ is the general solutions. If $f$ is assumed to be regular, for example continuous, the function $a$ has to be of the form $xmapsto c x$ with some constant $c$. Thus in particular holde true If $f$ is differentiable. In that case the condition on $f'$ results in the form of $f$ as given in the other answer.
answered Jul 20 at 10:54
Jens Schwaiger
1,112116
add a comment |Â
up vote
0
down vote
The equation itself may be solved without any regularity conditions by observing that $g$, $g(x):=f(x)-x$ satisfies the logarithmic equation $g(xy)=g(x)+g(y)$ for $gcolon (0,infty)tomathbbR$. Thus there is some additive function $acolonmathbbRtomathbbR$, i.e., $a$ satisfies $a(x+y)=a(x)+a(y)$ for all real $x,y$, such that $g(x)=a(ln x)$ for all $x>0$. Thus $f$ with $f(x):=a(ln x)+x$ is the general solutions. If $f$ is assumed to be regular, for example continuous, the function $a$ has to be of the form $xmapsto c x$ with some constant $c$. Thus in particular holde true If $f$ is differentiable. In that case the condition on $f'$ results in the form of $f$ as given in the other answer.
answered Jul 20 at 10:54
Jens Schwaiger
1,112116
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The equation itself may be solved without any regularity conditions by observing that $g$, $g(x):=f(x)-x$ satisfies the logarithmic equation $g(xy)=g(x)+g(y)$ for $gcolon (0,infty)tomathbbR$. Thus there is some additive function $acolonmathbbRtomathbbR$, i.e., $a$ satisfies $a(x+y)=a(x)+a(y)$ for all real $x,y$, such that $g(x)=a(ln x)$ for all $x>0$. Thus $f$ with $f(x):=a(ln x)+x$ is the general solutions. If $f$ is assumed to be regular, for example continuous, the function $a$ has to be of the form $xmapsto c x$ with some constant $c$. Thus in particular holde true If $f$ is differentiable. In that case the condition on $f'$ results in the form of $f$ as given in the other answer.
answered Jul 20 at 10:54
Jens Schwaiger
1,112116
The equation itself may be solved without any regularity conditions by observing that $g$, $g(x):=f(x)-x$ satisfies the logarithmic equation $g(xy)=g(x)+g(y)$ for $gcolon (0,infty)tomathbbR$. Thus there is some additive function $acolonmathbbRtomathbbR$, i.e., $a$ satisfies $a(x+y)=a(x)+a(y)$ for all real $x,y$, such that $g(x)=a(ln x)$ for all $x>0$. Thus $f$ with $f(x):=a(ln x)+x$ is the general solutions. If $f$ is assumed to be regular, for example continuous, the function $a$ has to be of the form $xmapsto c x$ with some constant $c$. Thus in particular holde true If $f$ is differentiable. In that case the condition on $f'$ results in the form of $f$ as given in the other answer.
answered Jul 20 at 10:54
Jens Schwaiger
1,112116
edited Jul 20 at 11:19
answered Jul 20 at 10:54
Jens Schwaiger
1,112116
answered Jul 20 at 10:54
Jens Schwaiger
1,112116
answered Jul 20 at 10:54
Jens Schwaiger
1,112116
1,112116
add a comment |Â
add a comment |Â
1
It's possible to solve the equation for $f$. Differentiate the equation wrt to $x$ holding $y$ fixed.
– Oliver Jones
Jul 20 at 6:20