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What is value of g'(2) when f(x) and g(x) are inverses and the following is given? [closed]

Finding a value so that a given f(x) is continuous at 0Find the derivative with respect to $x$ of the given combination at the given value of $x$.find the value of $f'(5)$ for given functionAlgebraically Deriving a function from a Table of ValuesLimits of the function given by the following graphFind the upper and lower Riemann sums $U(f,P)$ and $L(f,P)$ for discontinuous functionRecognizing Patterns in Alternating Signs Matricies and their InversesHow to prove that $(0,1]$ is equinumerous to $(0,1)$Why abstractly do left and right inverses coincide when $f$ is bijective?Set theory, functions and inverses

-4

0

$begingroup$

What is value of $g'(2)$ when $f(x)$ and $g(x)$ are inverses and
beginarray[rlrlrlrlrl]
f&f(2)&=1,quad f(1)&=2,quad f(5)&=7,quad &f(3)&=-2 \
&g(2)&=5,quad g(1)&=4,quad g(5)&=9,quad &g(3)&=-4 \
&f'(2)&=3,quad f'(1)&=6,quad f'(5)&=-1, &f'(3)&=-6\endarray

Original:

calculus functions inverse-function

share|cite|improve this question

edited Mar 28 at 15:29

MarianD

2,1281618

asked Mar 28 at 14:49

techlovertechlover

11

New contributor

techlover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

closed as off-topic by RRL, egreg, Michael Rybkin, Adrian Keister, max_zorn Mar 28 at 17:48

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, egreg, Michael Rybkin, Adrian Keister, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Please type the question out instead of just referencing it in a link, and show your efforts. Links can go dead.
  $endgroup$
  – YiFan
  Mar 28 at 14:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Answer is $frac 1 3.$
  $endgroup$
  – Dbchatto67
  Mar 28 at 15:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dbchatto67 Please can you explain ?
  $endgroup$
  – techlover
  Mar 28 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $f$ and $g$ are inverses of each other, then $f(g(x))=x$. Taking derivatives you get that $f'(g(x))cdot g'(x)=1$. Evaluating at $x=2$ you get that $f'(g(2))cdot g'(2)=1$. Since $g(2)=5$ and $f'(5)=-1$, you get that $-g'(2)=1$. Therefore, $g'(2)=-1$.
  $endgroup$
  – user647486
  Mar 28 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We know that if $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$ then $(g circ f)$ is differentiable at $x.$ This is known as chain rule for differentiation. According to your problem $g$ and $f$ are inverses of each other. So in this case $(g circ f) (x) = x, text for all x.$ Can you proceed now?
  $endgroup$
  – Dbchatto67
  Mar 28 at 15:11
  
  

  
  
  
  

 | 
show 3 more comments

-4

0

$begingroup$

What is value of $g'(2)$ when $f(x)$ and $g(x)$ are inverses and
beginarray[rlrlrlrlrl]
f&f(2)&=1,quad f(1)&=2,quad f(5)&=7,quad &f(3)&=-2 \
&g(2)&=5,quad g(1)&=4,quad g(5)&=9,quad &g(3)&=-4 \
&f'(2)&=3,quad f'(1)&=6,quad f'(5)&=-1, &f'(3)&=-6\endarray

Original:

calculus functions inverse-function

share|cite|improve this question

edited Mar 28 at 15:29

MarianD

2,1281618

asked Mar 28 at 14:49

techlovertechlover

11

New contributor

techlover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

closed as off-topic by RRL, egreg, Michael Rybkin, Adrian Keister, max_zorn Mar 28 at 17:48

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, egreg, Michael Rybkin, Adrian Keister, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Please type the question out instead of just referencing it in a link, and show your efforts. Links can go dead.
  $endgroup$
  – YiFan
  Mar 28 at 14:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Answer is $frac 1 3.$
  $endgroup$
  – Dbchatto67
  Mar 28 at 15:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dbchatto67 Please can you explain ?
  $endgroup$
  – techlover
  Mar 28 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $f$ and $g$ are inverses of each other, then $f(g(x))=x$. Taking derivatives you get that $f'(g(x))cdot g'(x)=1$. Evaluating at $x=2$ you get that $f'(g(2))cdot g'(2)=1$. Since $g(2)=5$ and $f'(5)=-1$, you get that $-g'(2)=1$. Therefore, $g'(2)=-1$.
  $endgroup$
  – user647486
  Mar 28 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We know that if $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$ then $(g circ f)$ is differentiable at $x.$ This is known as chain rule for differentiation. According to your problem $g$ and $f$ are inverses of each other. So in this case $(g circ f) (x) = x, text for all x.$ Can you proceed now?
  $endgroup$
  – Dbchatto67
  Mar 28 at 15:11
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

What is value of $g'(2)$ when $f(x)$ and $g(x)$ are inverses and
beginarray[rlrlrlrlrl]
f&f(2)&=1,quad f(1)&=2,quad f(5)&=7,quad &f(3)&=-2 \
&g(2)&=5,quad g(1)&=4,quad g(5)&=9,quad &g(3)&=-4 \
&f'(2)&=3,quad f'(1)&=6,quad f'(5)&=-1, &f'(3)&=-6\endarray

Original:

calculus functions inverse-function

share|cite|improve this question

edited Mar 28 at 15:29

MarianD

2,1281618

asked Mar 28 at 14:49

techlovertechlover

11

New contributor

techlover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question

edited Mar 28 at 15:29

MarianD

2,1281618

asked Mar 28 at 14:49

techlovertechlover

11

New contributor

techlover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question

edited Mar 28 at 15:29

MarianD

2,1281618

asked Mar 28 at 14:49

techlovertechlover

11

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edited Mar 28 at 15:29

MarianD

2,1281618

edited Mar 28 at 15:29

MarianD

2,1281618

asked Mar 28 at 14:49

techlovertechlover

11

New contributor

techlover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Mar 28 at 14:49

techlovertechlover

11

4 Answers
4

active

oldest

votes

0

$begingroup$

Hint $:$ Chain rule for differentiation.

We know that if $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$ then $(g circ f)$ is differentiable at $x$ and $(g circ f)'(x) = g'(f(x)) cdot f'(x).$ This is known as chain rule for differentiation. According to your problem $g$ and $f$ are inverses of each other. So in this case $(g circ f) (x) = x, text for all x.$ Here $f(1) = 2.$ Since $g$ is differentiable at $2$ so $(g circ f)$ is differentiable at $1.$ So by using chain rule for differentiation we have

$1 = (g circ f)' (1) = g'(f(1)) cdot f'(1) = 6 cdot g'(2).$ Therefore $$g'(2) = frac 1 6.$$

share|cite|improve this answer

edited Mar 28 at 15:16

answered Mar 28 at 15:01

Dbchatto67Dbchatto67

2,445522

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  f(g(x)) =x ; f'(g(x)) * g'(x) = 1; Am i doing it correctly ?
  $endgroup$
  – techlover
  Mar 28 at 15:08
  
  

  
  
  
  

add a comment | 

0

$begingroup$

How can $g$ and $f$ be inverses if $f(2)=1$ and $g(1)=4$ ?

share|cite|improve this answer

answered Mar 28 at 15:49

Antoine FalckAntoine Falck

1

New contributor

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$endgroup$

add a comment | 

0

$begingroup$

$$g(f(x))=ximplies g'(f(x))cdot f'(x)=1$$

or, for $x=1$,

$$g'(2)cdot f'(1)=1iff g'(2)=frac16$$

Similarly, $$f(g(x))=xiff g'(2)=-1neqfrac16$$ for $x=2$.

This contradiction means that such a pair of functions $(f,g)$ does not even exist.

share|cite|improve this answer

edited Mar 28 at 16:03

answered Mar 28 at 15:12

st.mathst.math

1,05615

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What's wrong with :- f(g(x)) = x or, f'(g(x)) * g'(x) = 1 for x = 2, f'(g(2)) * g'(2) = 1 or, f'(5) * g'(2) = 1 or, -1 * g'(2) = 1 or, g'(2) = -1
  $endgroup$
  – techlover
  Mar 28 at 15:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right; the fact that there are different solutions ($-1$ and $1/6$) implies that such a pair of functions $f,g$ does not exist. Edited my answer to include this fact.
  $endgroup$
  – st.math
  Mar 28 at 16:00
  
  

  
  
  
  

add a comment | 

0

$begingroup$

The rule for derivative of the inverse function $g$ is as simple as

$$g'(y) = 1over f'(x),quad textwhere y = f(x).tag 1$$

From your table: $2 = f(1), $ so substituting $ y=2, x=1$ into $(1)$ you will obtain

$$g'(2) = 1over f'(1) tag 2$$

But - from your table, too - $f'(1) = 6, $ so substituting it into $(2)$ you reach your goal:

$$colorredg'(2) = 1over 6$$

share|cite|improve this answer

answered Mar 28 at 16:06

MarianDMarianD

2,1281618

$endgroup$

add a comment | 

0

$begingroup$

Hint $:$ Chain rule for differentiation.

We know that if $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$ then $(g circ f)$ is differentiable at $x$ and $(g circ f)'(x) = g'(f(x)) cdot f'(x).$ This is known as chain rule for differentiation. According to your problem $g$ and $f$ are inverses of each other. So in this case $(g circ f) (x) = x, text for all x.$ Here $f(1) = 2.$ Since $g$ is differentiable at $2$ so $(g circ f)$ is differentiable at $1.$ So by using chain rule for differentiation we have

$1 = (g circ f)' (1) = g'(f(1)) cdot f'(1) = 6 cdot g'(2).$ Therefore $$g'(2) = frac 1 6.$$

share|cite|improve this answer

edited Mar 28 at 15:16

answered Mar 28 at 15:01

Dbchatto67Dbchatto67

2,445522

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  f(g(x)) =x ; f'(g(x)) * g'(x) = 1; Am i doing it correctly ?
  $endgroup$
  – techlover
  Mar 28 at 15:08
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint $:$ Chain rule for differentiation.

We know that if $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$ then $(g circ f)$ is differentiable at $x$ and $(g circ f)'(x) = g'(f(x)) cdot f'(x).$ This is known as chain rule for differentiation. According to your problem $g$ and $f$ are inverses of each other. So in this case $(g circ f) (x) = x, text for all x.$ Here $f(1) = 2.$ Since $g$ is differentiable at $2$ so $(g circ f)$ is differentiable at $1.$ So by using chain rule for differentiation we have

$1 = (g circ f)' (1) = g'(f(1)) cdot f'(1) = 6 cdot g'(2).$ Therefore $$g'(2) = frac 1 6.$$

share|cite|improve this answer

edited Mar 28 at 15:16

answered Mar 28 at 15:01

Dbchatto67Dbchatto67

2,445522

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  f(g(x)) =x ; f'(g(x)) * g'(x) = 1; Am i doing it correctly ?
  $endgroup$
  – techlover
  Mar 28 at 15:08
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint $:$ Chain rule for differentiation.

We know that if $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$ then $(g circ f)$ is differentiable at $x$ and $(g circ f)'(x) = g'(f(x)) cdot f'(x).$ This is known as chain rule for differentiation. According to your problem $g$ and $f$ are inverses of each other. So in this case $(g circ f) (x) = x, text for all x.$ Here $f(1) = 2.$ Since $g$ is differentiable at $2$ so $(g circ f)$ is differentiable at $1.$ So by using chain rule for differentiation we have

$1 = (g circ f)' (1) = g'(f(1)) cdot f'(1) = 6 cdot g'(2).$ Therefore $$g'(2) = frac 1 6.$$

share|cite|improve this answer

edited Mar 28 at 15:16

answered Mar 28 at 15:01

Dbchatto67Dbchatto67

2,445522

$endgroup$

share|cite|improve this answer

edited Mar 28 at 15:16

answered Mar 28 at 15:01

Dbchatto67Dbchatto67

2,445522

answered Mar 28 at 15:01

Dbchatto67Dbchatto67

2,445522

answered Mar 28 at 15:01

Dbchatto67Dbchatto67

2,445522

0

$begingroup$

How can $g$ and $f$ be inverses if $f(2)=1$ and $g(1)=4$ ?

share|cite|improve this answer

answered Mar 28 at 15:49

Antoine FalckAntoine Falck

1

New contributor

Antoine Falck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

How can $g$ and $f$ be inverses if $f(2)=1$ and $g(1)=4$ ?

share|cite|improve this answer

answered Mar 28 at 15:49

Antoine FalckAntoine Falck

1

New contributor

Antoine Falck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

How can $g$ and $f$ be inverses if $f(2)=1$ and $g(1)=4$ ?

share|cite|improve this answer

answered Mar 28 at 15:49

Antoine FalckAntoine Falck

1

New contributor

Antoine Falck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered Mar 28 at 15:49

Antoine FalckAntoine Falck

1

New contributor

Antoine Falck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 28 at 15:49

Antoine FalckAntoine Falck

1

New contributor

Antoine Falck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 28 at 15:49

Antoine FalckAntoine Falck

1

0

$begingroup$

$$g(f(x))=ximplies g'(f(x))cdot f'(x)=1$$

or, for $x=1$,

$$g'(2)cdot f'(1)=1iff g'(2)=frac16$$

Similarly, $$f(g(x))=xiff g'(2)=-1neqfrac16$$ for $x=2$.

This contradiction means that such a pair of functions $(f,g)$ does not even exist.

share|cite|improve this answer

edited Mar 28 at 16:03

answered Mar 28 at 15:12

st.mathst.math

1,05615

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What's wrong with :- f(g(x)) = x or, f'(g(x)) * g'(x) = 1 for x = 2, f'(g(2)) * g'(2) = 1 or, f'(5) * g'(2) = 1 or, -1 * g'(2) = 1 or, g'(2) = -1
  $endgroup$
  – techlover
  Mar 28 at 15:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right; the fact that there are different solutions ($-1$ and $1/6$) implies that such a pair of functions $f,g$ does not exist. Edited my answer to include this fact.
  $endgroup$
  – st.math
  Mar 28 at 16:00
  
  

  
  
  
  

add a comment | 

0

$begingroup$

$$g(f(x))=ximplies g'(f(x))cdot f'(x)=1$$

or, for $x=1$,

$$g'(2)cdot f'(1)=1iff g'(2)=frac16$$

Similarly, $$f(g(x))=xiff g'(2)=-1neqfrac16$$ for $x=2$.

This contradiction means that such a pair of functions $(f,g)$ does not even exist.

share|cite|improve this answer

edited Mar 28 at 16:03

answered Mar 28 at 15:12

st.mathst.math

1,05615

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What's wrong with :- f(g(x)) = x or, f'(g(x)) * g'(x) = 1 for x = 2, f'(g(2)) * g'(2) = 1 or, f'(5) * g'(2) = 1 or, -1 * g'(2) = 1 or, g'(2) = -1
  $endgroup$
  – techlover
  Mar 28 at 15:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right; the fact that there are different solutions ($-1$ and $1/6$) implies that such a pair of functions $f,g$ does not exist. Edited my answer to include this fact.
  $endgroup$
  – st.math
  Mar 28 at 16:00
  
  

  
  
  
  

add a comment | 

$begingroup$

$$g(f(x))=ximplies g'(f(x))cdot f'(x)=1$$

or, for $x=1$,

$$g'(2)cdot f'(1)=1iff g'(2)=frac16$$

Similarly, $$f(g(x))=xiff g'(2)=-1neqfrac16$$ for $x=2$.

This contradiction means that such a pair of functions $(f,g)$ does not even exist.

share|cite|improve this answer

edited Mar 28 at 16:03

answered Mar 28 at 15:12

st.mathst.math

1,05615

$endgroup$

share|cite|improve this answer

edited Mar 28 at 16:03

answered Mar 28 at 15:12

st.mathst.math

1,05615

answered Mar 28 at 15:12

st.mathst.math

1,05615

answered Mar 28 at 15:12

st.mathst.math

1,05615

0

$begingroup$

The rule for derivative of the inverse function $g$ is as simple as

$$g'(y) = 1over f'(x),quad textwhere y = f(x).tag 1$$

From your table: $2 = f(1), $ so substituting $ y=2, x=1$ into $(1)$ you will obtain

$$g'(2) = 1over f'(1) tag 2$$

But - from your table, too - $f'(1) = 6, $ so substituting it into $(2)$ you reach your goal:

$$colorredg'(2) = 1over 6$$

share|cite|improve this answer

answered Mar 28 at 16:06

MarianDMarianD

2,1281618

$endgroup$

add a comment | 

0

$begingroup$

The rule for derivative of the inverse function $g$ is as simple as

$$g'(y) = 1over f'(x),quad textwhere y = f(x).tag 1$$

From your table: $2 = f(1), $ so substituting $ y=2, x=1$ into $(1)$ you will obtain

$$g'(2) = 1over f'(1) tag 2$$

But - from your table, too - $f'(1) = 6, $ so substituting it into $(2)$ you reach your goal:

$$colorredg'(2) = 1over 6$$

share|cite|improve this answer

answered Mar 28 at 16:06

MarianDMarianD

2,1281618

$endgroup$

add a comment | 

$begingroup$

The rule for derivative of the inverse function $g$ is as simple as

$$g'(y) = 1over f'(x),quad textwhere y = f(x).tag 1$$

From your table: $2 = f(1), $ so substituting $ y=2, x=1$ into $(1)$ you will obtain

$$g'(2) = 1over f'(1) tag 2$$

But - from your table, too - $f'(1) = 6, $ so substituting it into $(2)$ you reach your goal:

$$colorredg'(2) = 1over 6$$

share|cite|improve this answer

answered Mar 28 at 16:06

MarianDMarianD

2,1281618

$endgroup$

share|cite|improve this answer

answered Mar 28 at 16:06

MarianDMarianD

2,1281618

answered Mar 28 at 16:06

MarianDMarianD

2,1281618

answered Mar 28 at 16:06

MarianDMarianD

2,1281618